Problem 15.1 In Active Example 15.1, what is the velocity of the container when it has reached the position s = 2m?


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1 Problem 5. In Active Example 5., what is the velocity of the container when it has reached the position s = m? A s The 8kg container A starts from rest at s =. The horizontal force (in newtons) is F = 7 5s. The coefficient of kinetic friction is µ k =.6. U = T T (7 5s.6[8(9.8)])ds = (8 kg)v 7() (5)() (.6[8(9.8)][]) = 9v v =.4 m/s. Problem 5. The mass of the Sikorsky UH6A helicopter is 93 kg. It takes off vertically with its rotor exerting a constant upward thrust of kn. Use the principle of work and energy to determine how far it has risen when its velocity is 6 m/s. Strategy: Be sure to draw the freebody diagram of the helicopter. U = T T [( 93[9.8])N]h = (93 kg)(6 m/s) h = 8.6 m. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 79
2 Problem 5.3 The N box is at rest on the horizontal surface when the constant force F = 5 N is applied. The coefficient of kinetic friction between the box and the surface is µ k =.. Determine how fast the box is moving when it has moved m from its initial position by applying Newton s second law; by applying the principle of work and energy. F The equations of motion can be used to find the acceleration F x : F f = W g a, F y : N W =, f = µ k N Solving we have ( ) ) F 5N a = g W µ k = ( )( N. 9.8m/s =.49 m/s Now we integrate to find the velocity at the new position a = v dv v m ds vdv = ads v = a( m) = (.49 m/s )( m) v =. 4 m/s Using the principle of work and energy we have (recognizing that N = W ) U = T T (F µ k N)d = ( ) W v g ( ) ) F 5N v = g W µ k d = ( 9.8 m/s )( N. ( m) v =.4 m/s Problem 5.4 At the instant shown, the 3N box is moving up the smooth inclined surface at m/s. The constant force F = 5 N. How fast will the box be moving when it has moved m up the surface from its present position? F U = T T = Solving we find [(5 N) cos (3 N) sin ]( m) ( ) 3 N v ( ) 3 N ( m/s) 9.8 m/s 9.8 m/s v =.55 m/s. 8 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
3 Problem 5.5 The.45kg soccer ball is m above the ground when it is kicked straight upward at m/s. By using the principle of work and energy, determine: how high above the ground the ball goes, the magnitude of the ball s velocity when it falls back to a height of m above the ground, (c) the magnitude of the ball s velocity immediately before it hits the ground. m/s m Find the height above the ground mg( m h) = mv, h = v g + m= ( m/s) (9.8 m/s + m= 6. m ) When the ball returns to the same level, the velocity must be equal to the initial velocity (but now it is moving downward) because the net work is zero v = m/s (c) The velocity just before it hits the ground mg( m) = mv mv v = v + g( m) = ( m/s) + (9.8 m/s )( m) v =.9 m/s. h = 6. m, v =. m/s, (c) v =.9 m/s. Problem 5.6 Assume that the soccer ball in Problem 5.5 is stationary the instant before it is kicked upward at m/s. The duration of the kick is. s. What average power is transferred to the ball during the kick? U = (.45 kg)( m/s) = 3.4 Nm Power = U t = 3.4 Nm. s =.6 kw c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 8
4 Problem 5.7 The N drag racer starts from rest and travels a quarterkilometre course. It completes the cours in 4.54 seconds and crosses the finish line traveling at km/h. How much work is done on the car as it travels the course? Assume that the horizontal force exerted on the car is constant and use the principle of work and energy to determine it. e The work is equal to the change in kinetic energy. U = mv = ( )[ ( N )] (35.77 km/h) 9.8 m/s 36 U = Nm The work is equal to the force times the distance U = Fd F = U d = Nm 4 ( ) = 3339 N m F = 3339 N Problem 5.8 The N drag racer starts from rest and travels a quarterkilometre course. It completes the course in 4.54 seconds and crosses the finish line traveling at km/h. Assume that the horizontal force exerted on the car is constant. Determine the maximum power and the average power transferred to the car as it travels the quarterkilometre course. From problem 5.7 we know that the force is The maximum power occurs when the car has reached its maximum velocity ( ) P = Fv= ( 3339 N)(35.77 km/h) = 3. 5 N m/s. 36 The average power is equal to the change in kinetic energy divided by the time. ( )[ ( )] N (35.77 km/h) P ave = mv 9.8 m/s 36 = t 4.54 s 5 =. 845 Nm/s Nm/s, Nm/s. 8 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
5 Problem 5.9 As the 3,N airplane takes off, the tangential component of force exerted on it by its engines is F t = 45, N. Neglecting other forces on the airplane, use the principle of work and energy to determine how much runway is required for its velocity to reach km/h. U = mv Fd = mv d = mv F ( )[ ( 3, N )] ( km/h) 9.8 m/s 36 d = = m. (45, N) d = m Problem 5. As the 3,N airplane takes off, the tangential component of force exerted on it by its engines is F t = 45, N. Neglecting other forces on the airplane, determine the maximum power and the average power transferred to the airplane as its velocity increases from zero to km/h. The maximum power occurs when the velocity is a maximum [ ] P = Fv = (45, N) km/h =.5 6 N m/s. 36 To find the average power we need to know the time that it takes to reach full speed a = F m = ( 45, N ) = 3.8 m/s 3, N 9.8 m/s v = at t = v km/h a = 36 = 4.3 s. 3.8 m/s Now, the average power is the change in kinetic energy divided by the time ( )( ) 3, N P ave = mv km/h 9.8 m/s 36 6 = =.5 Nm/s. t 4.3 s.5 6 Nm/s,.5 6 Nm/s. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 83
6 Problem 5. The 3,N airplane takes off from rest in the position s =. The total tangential force exerted on it by its engines and aerodynamic drag (in Newtons) is given as a function of its position s by F t = 45, 5.s. Use the principle of work and energy to determine how fast the airplane is traveling when its position is s = 95 m. 95 U = (45, 5.s)ds = (45, )(95) (5.)(95) = Nm U = mv = ( ) 3, N v 9.8 m/s Solving, we find v = 57.4 m/s. Problem 5. The spring (k = N/m) is unstretched when s =. The 5kg cart is moved to the position s = m and released from rest. What is the magnitude of its velocity when it is in the position s =? k s First we calculate the work done by the spring and by gravity U = ( ks + mg sin ) ds m = k( m) + mg sin (m) = ( N/m)( m) + (5 kg)(9.8 m/s ) sin (m) = 6.8 Nm. Now using work and energy U = mv v = U m = (6.8 Nm) = 3.7 m/s. 5kg v = 3.7 m/s. 84 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
7 Problem 5.3 The spring (k = N/m) is unstretched when s =. The 5kg cart is moved to the position s = m and released from rest. What maximum distance down the sloped surface does the cart move relative to its initial position? k s The cart starts from a position of rest, and when it reaches the maximum position, it is again at rest. Therefore, the total work must be zero. s U = ( ks + mg sin ) ds m = k(s [ m] ) + mg sin (s [ m]) = ( N/m)(s [ m] ) + (5 kg)(9.8 m/s ) sin (s [ m]) = This is a quadratic equation that has the two solutions s = m,s =.68 m. The distance relative to the initial is s = s + m. s = 3.68 m. Problem 5.4 The force exerted on a car by a prototype crash barrier as the barrier crushes is F = (s + 4s 3 ) N, where s is the distance in metre from the initial contact. The effective length of the barrier is 8 m. How fast can a 5N car be moving and be brought to rest within the effective length of the barrier? s The barrier can provide a maximum amount of work given by 8 U = (s + 4s 3 ) ds = ()(8) 4 (4)(8)4 =.7 6 Nm. Using work and energy, we have U = mv.7 6 m = ( ) 5 N N v 9.8 m/s Solving for the velocity, we find v = 64.8 m/s. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 85
8 Problem 5.5 A 5N car hits the crash barrier at 8 km/h and is brought to rest in. seconds. What average power is transferred from the car during the impact? s The average power is equal to the change in kinetic energy divided by the time ( )( ) 5 N 8 km/h P = mv 9.8 m/s 36 = =.4 6 N m/s. t. s P =.4 6 Nm/s. Problem 5.6 A group of engineering students constructs a sunpowered car and tests it on a circular track with a m radius. The car, with a weight of 46 N including its occupant, starts from rest. The total tangential component of force on the car is F t = 3.s N, where s is the distance (in ft) the car travels along the track from the position where it starts. Determine the work done on the car when it has gone a distance s = m. Determine the magnitude of the total horizontal force exerted on the car s tires by the road when it is at the position s = m. m U = [3. s] Nds = 6 Nm 6 Nm = ( ) 46 N v 9.8 m/s v = 9.6 m/s F t = [3.()] = 6N ( ) F n = m v 46 N ( ) ρ = 9.6 m/s 9.8 m/s m = 4.3 N F = F t + F n = (6N) + (4.3 N) = 7.39 N 86 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
9 Problem 5.7 At the instant shown, the 6N vaulter s center of mass is 8.5 m above the ground, and the vertical component of his velocity is 4 m/s. As his pole straightens, it exerts a vertical force on the vaulter of magnitude 8 +.8y N, where y is the vertical position of his center of mass relative to its position at the instant shown. This force is exerted on him from y = toy = 4 m, when he releases the pole. What is the maximum height above the ground reached by the vaulter s center of mass? The work done on him by the pole is 4 U pole = (8 +.8 y )dy = 8(4) +.8 (4)3 3 = 78 N m. Let y max be his maximum height above the ground. The work done by his weight from the instant shown to the maximum height is 6(y max 8.5) = U weight,oru weight + U pole = mv / mv / 78 6(y max 8.5) = ( ) 6 (4). 9.8 Solving, y max = 4. m Problem 5.8 The springs (k = 5 N/ cm) are unstretched when s =. The 5N weight is released from rest in the position s =. When the weight has fallen cm, how much work has been done on it by each spring? What is the magnitude of the velocity of the weight when it has fallen cm? k k s The work done by each spring cm U = ksds = (5 N/cm)(cm) =.5 Ncm. The velocity is found from the workenergy equation. The total work includes the work done by both springs and by gravity U = (5 N)( cm) (.5 Ncm) = 5 Ncm. U = mv = ( ) 5 N v 9.8 m/s Solving for the velocity we find v =.3 m/s..5 Ncm,.3 m/s. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 87
10 Problem 5.9 The coefficients of friction between the 6kg crate and the ramp are µ s =.3 and µ k =.8. What tension T must the winch exert to start the crate moving up the ramp? If the tension remains at the value T after the crate starts sliding, what total work is done on the crate as it slides a distance s = 3 m up the ramp, and what is the resulting velocity of the crate? 8 s The tension is T = W sin θ + µ s N, from which T = mg(sin θ + µ s cos θ) = 93.9 N. The work done on the crate by (nonfriction) external forces is mg µ s N N T T mg N µ k N 3 3 U weight = T ds (mg sin θ)ds = 93.9(3) 455. = Nm. The work done on the crate by friction is 3 U f = ( µ k N)ds = 3µ k mg cos θ = 53.9 Nm. From the principle of work and energy is U weight + U f = mv, from which 6(T mg(sin θ + µ k cos θ)) v = m v =.6 m/s Problem 5. In Problem 5.9, if the winch exerts a tension T = T ( +.s) after the crate starts sliding, what total work is done on the crate as it slides a distance s = 3 m up the ramp, and what is the resulting velocity of the crate? The work done on the crate is From the principle of work and energy, U = mv, from which U = Tds (mg sin θ)ds µ k (mg cos θ)ds, from which v = U m =.5 m/s U = T [( s +.5s )] 3 (mg sin θ)(3) µ k(mg cos θ)(3). From the solution to Problem 5.9, T = 93.9 Nm, from which the total work done is U = = Nm. 88 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
11 Problem 5. The mmdiameter gas gun is evacuated on the right of the 8kg projectile. On the left of the projectile, the tube contains gas with pressure p = 5 Pa (N/m ). The force F is slowly increased, moving the projectile.5 m to the left from the position shown. The force is then removed and the projectile accelerates to the right. If you neglect friction and assume that the pressure of the gas is related to its volume by pv = constant, what is the velocity of the projectile when it has returned to its original position? m Gas Projectile F The constant is K = pv = 5 ()(.) π = 34.6 Nm. The force is F = pa. The volume is V = As, from which the pressure varies as the inverse distance: p = K, from which As F = K s. From the principle of work and energy, the work done by the gas is equal to the gain in kinetic energy: K ln() = mv, and v = K m ln(), The work done by the gas is K U = Fds= ds = [K ln(s)] x = K ln(). v = K m ln() = 3.33 m/s Note: The argument of ln() is dimensionless, since it is ratio of two distances. Problem 5. In Problem 5., if you assume that the pressure of the gas is related to its volume by pv = constant while it is compressed (an isothermal process) and by pv.4 = constant while it is expanding (an isentropic process), what is the velocity of the projectile when it has returned to its original position? The isothermal constant is K = 34.6 Nm from the solution to Problem 5.. The pressure at the leftmost position is p = K A(.5) = 5 N/m. The isentropic expansion constant is K e = pv.4 = ( 5 )(A.4 )(.5.4 ) = Nm The pressure during expansion is From the principle of work and energy, the work done is equal to the gain in kinetic energy, Fds=.5 mv, from which the velocity is (9.8) v = =.8 m/s. m p = K e (As).4 = K e A.4 s.4. The force is F = pa = K e A.4 s.4. The work done by the gas during expansion is.. [ s U = Fds= K e A.4 s.4 ds = K e A.4.4 ] = 9.8 Nm. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 89
12 Problem 5.3 In Example 5., suppose that the angle between the inclined surface and the horizontal is increased from to 3. What is the magnitude of the velocity of the crates when they have moved 4 mm? v A Doing work energy for the system B.4 (m A g sin 3 µ k m A g cos 3 + m B g)ds = (m A + m B )v v [4 sin 3 (.5)(4) cos 3 + 3](9.8)(.4) = (7)v Solving for the velocity we find v =.4 m/s. Problem 5.4 The system is released from rest. The 4kg mass slides on the smooth horizontal surface. By using the principle of work and energy, determine the magnitude of the velocity of the masses when the kg mass has fallen m. 4 kg kg Write workenergy for system U = ( kg)(9.8 m/s )( m) = (4 kg)v v = 4.4 m/s Problem 5.5 Solve Problem 5.4 if the coefficient of kinetic friction between the 4kg mass and the horizontal surface is µ k =.4. (4 kg)(9.8 m/s ) Fy : N (4 kg)(9.8 m/s ) = N = 39.4 N Write workenergy for system U = [( kg)(9.8 m/s ).4(39.4 N)]( m) = 8.5 Nm.4 N T 8.5 Nm = (4 kg)v v = 3.88 m/s N 9 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
13 Problem 5.6 Each box weighs 5 N and the inclined surfaces are smooth. The system is released from rest. Determine the magnitude of the velocities of the boxes when they have moved m Write workenergy for the system U = (5 N sin 45 )( m) (5 N sin 3 )( m) =.36 Nm.36 Nm = ( ) N v v =.43 m/s 9.8 m/s Problem 5.7 Solve Problem 5.6 if the coefficient of kinetic friction between the boxes and the inclined surfaces is µ k =.5. 5 N 5 N F : N (5 N) sin 45 = F : N (5 N) cos 3 =.5 N N = 35.4 N, N = 43.3 N Workenergy for the system U = (5 N sin 45 )( m) (.5)(35.4 N)( m) N N.5 N (5 N sin 3)( m) (.5)(43.3 N)( m) = 6.4 Nm 6.4 Nm = ( ) N v v =. m/s 9.8 m/s c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 9
14 Problem 5.8 The masses of the three blocks are m A = 4 kg, m B = 6 kg, and m C = kg. Neglect the mass of the bar holding C in place. Friction is negligible. By applying the principle of work and energy to A and B individually, determine the magnitude of their velocity when they have moved 5 mm. B C A 45 Denote b =.5 m. Since the pulley is onetoone, denote v A = v B =v. The principle of work and energy for weight A is b (m A g sin θ T)ds = m Av, b and for weight B (T m B g sin θ)ds = m Bv. T m A g T m B g N AB N AB N BC Add the two equations: N A (m A m B )gb sin θ = (m A + m B )v. (ma m Solve: B )gb sin θ v A = v B = =.7 m/s (m A + m B ) Problem 5.9 Solve Problem 5.8 by applying the principle of work and energy to the system consisting of A, B, the cable connecting them, and the pulley. Choose a coordinate system with the origin at the pulley axis and the positive x axis parallel to the inclined surface. Since the pulley is onetoone, x A = x B. Differentiate to obtain v A = v B. Denote b =.5 m. From the principle of work and energy the work done by the external forces on the complete system is equal to the gain in kinetic energy, xa xb m A g sin θds+ m B g sin θds= m AvA + m BvB, m B g N BC from which m A g (m B m A )gb sin θ = (m A + m B )v A N A (ma m and B ) v A = v B = gb sin θ =.7 m/s. (m A + m B ) 9 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
15 Problem 5.3 The masses of the three blocks are m A = 4 kg, m B = 6 kg, and m C = kg. The coefficient of kinetic friction between all surfaces is µ k =.. Determine the magnitude of the velocity of blocks A and B when they have moved 5 mm. (See Example 5.3.) We will apply the principles of work energy to blocks A and B individually in order to properly account for the work done by internal friction forces. A B C b (m A g sin θ T µ k N A µ k N AB ) ds = m Av, 45 b (T m B g sin θ µ k N BC µ k N AB ) ds = m Bv. Adding the two equations, we get ([m A m B ]g sin θ µ k [N A + N AB + N BC ])b = (m A + m B )v The normal forces are N A = (m A + m B + m C )g cos θ, N AB = (m B + m C )g cos θ, N BC = (m C )g cos θ. Substitute and solve v =.4 m/s. Problem 5.3 In Example 5.5, suppose that the skier is moving at m/s when he is in position. Determine the horizontal component of his velocity when he reaches position, m below position. 3 U = mg(y y ) = mv mv Solving for v we find m(9.8)( ) = mv m() v = 8. m/s. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 93
16 Problem 5.3 Suppose that you stand at the edge of a 6 m cliff and throw rocks at 9. m/s in the three directions shown. Neglecting aerodynamic drag, use the principle of work and energy to determine the magnitude of the velocity of the rock just before it hits the ground in each case. 3 3 (c) 6 m U = m( 9.8 m/s )( 6 m) = mv m( 9. m/s) v = 35.7 m/s Note that the answer does not depend on the initial angle. Problem 5.33 The 3kg box is sliding down the smooth surface at m/s when it is in position. Determine the magnitude of the box s velocity at position in each case. m The work done by the weight is the same in both cases. U = m(9.8 m/s )( m) = mv m( m/s) 6 4 v = 6.34 m/s Problem 5.34 Solve Problem 5.33 if the coefficient of kinetic friction between the box and the inclined surface is µ k =.. The work done by the weight is the same, however, the work done by friction is different. U = m(9.8 m/s )( m) (.)[m(9.8 m/s ) cos 6 ] [ ] m sin 6 U = mv m( m/s) v = 5.98 m/s U = m(9.8 m/s )( m) (.)[m(9.8 m/s ) cos 4 ] [ ] m sin 4 U = mv m( m/s) v = 5.56 m/s Problem 5.35 In case, a 5N ball is released from rest at position and falls to position. In case, the ball is released from rest at position and swings to position. For each case, use the principle of work and energy to determine the magnitude of the ball s velocity at position. (In case, notice that the force exerted on the ball by the string is perpendicular to the ball s path.) m the same. The work is independent of the path, so both cases are U = m( 9.8m/s )( m) = mv v = 6.6 m/s 94 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
17 Problem 5.36 The kg ball is released from rest in position with the string horizontal. The length of the string is L = m. What is the magnitude of the ball s velocity when it is in position? 4 L kg L sin α U = mgj dsj U = mg( L sin α) = ()(9.8)() sin 4 α m L = m L SIN α U =.6 Nm U = mv mv v ()v =.6 v = 3.55 m/s Problem 5.37 The kg ball is released from rest in position with the string horizontal. The length of the string is L = m. What is the tension in the string when the ball is in position? Strategy: Draw the freebody diagram of the ball when it is in position and write Newton s second law in terms of normal and tangential components. m = kg Fr : T + mg cos 5 = mv /L 4 From Problem 5.36, T v = 3.55 m/s T = mg cos 5 + mv /L e r e θ T = 37.8 N 5 mg c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 95
18 Problem 5.38 The 4N wrecker s ball swings at the end of a 5m cable. If the magnitude of the ball s velocity at position is 4 m/s, what is the magnitude of its velocity just before it hits the wall at position? U = (4 N)( 5 m sin 95 [ 5 m sin 65 ]) U = ( ) 4 N (v [4 m/s] ) 9.8 m/s v = 7.75 m/s Problem 5.39 The 4N wrecker s ball swings at the end of a 5m cable. If the magnitude of the ball s velocity at position is 4 m/s, what is the maximum tension in the cable as the ball swings from position to position? From the solution to Problem 5.37, the tension in the cable is T = mg sin α + m v. From the solution to Problem 5.38, L v = gl[sin α sin 65 ](65 α 95 ), from which T = 3 mg sin α mg sin 65. The maximum tension occurs when sin α is a maximum in the interval (65 α 95 ), from which T = 3 mg sin 9 mg sin 65 = N. 96 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
19 Problem 5.4 A stunt driver wants to drive a car through the circular loop of radius R = 5 m. Determine the minimum velocity v at which the car can enter the loop and coast through without losing contact with the track. What is the car s velocity at the top of the loop? R First, let us find V T V TOP = V T Fn : N + mg = mv T /R For minimum velocity, N mg = mv T /R ds = dxi + dyj R V T = Rg = 7. m/s Now find V using workenergy V U T = mgj (dxi + dyj) N U T = mg dy = mgy e t U T = 98.m (Nm) Also, U T = mv T mv = 98.m (Nm) e n mg Solving for V (V T = 7. m/s) V = V T + (98.)() V = 5.68 m/s Problem 5.4 The kg collar starts from rest at position and slides down the smooth rigid wire. The yaxis points upward. What is the magnitude of the velocity of the collar when it reaches position? y (5, 5, ) m kg x z (3,, 3) m The work done by the weight is U weight = mgh, where h = y y = 5 ( ) = 6 m. From the principle of work and energy, mgh = mv, from which v = gh =.85 m/s c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 97
20 Problem 5.4 The 4N collar slides down the smooth rigid wire from position to position. When it reaches position, the magnitude of its velocity is 4 m/s. What was the magnitude of its velocity at position? (, 6, 4) m 4 N y x z (4nN,, 4) m U = (4 N)(6 [ ]) = ( ) 4N m /s ([4 m/s] v 9.8 m ) v =.9 m/s Problem 5.43 The forces acting on the 5 kn airplane are the thrust T and drag D, which are parallel to the airplane s path, the lift L, which is perpendicular to the path, and the weight W. The airplane climbs from an altitude of 94 m to an altitude of 348 m. During the climb, the magnitude of its velocity decreases from 44 m/s to 83 m/s. What work is done on the airplane by its lift during the climb? What work is done by the thrust and drag combined? D L W T The work due to the lift L is zero since it acts perpendicular to the motion. U = U L+D ( 5 N)( ) m U = ( ) 5 N ([83 m/s] [44 m/s] ) 9.8 m/s U L+D =.7 7 Nm 98 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
21 Problem 5.44 The.7 kn car is traveling 64.4 km/h at position. If the combined effect of the aerodynamic drag on the car and the tangential force exerted on its wheels by the road is that they exert no net tangential force on the car, what is the magnitude of the car s velocity at position? m 3.5 m 3 The initial velocity is v = km/h = 7. 9 m/s The change in elevation of the car is h = 36.6( cos 3 ) + 3.5( cos 3 ) = 67.( cos 3 ) = 8.98 m From the principle of work and energy the work done is equal to the gain in kinetic energy: U gravity = from which ( ) W v g ( ) W v g, g( ) v = =.96 m/ s W The initial kinetic energy is ( ) W v g = N The work done by gravity is h U gravity = ( W)ds = Wh = 7(h) = Nm. Problem 5.45 The.7 kn car is traveling 64.4 km/h at position. If the combined effect of aerodynamic drag on the car and the tangential force exerted on its wheels by the road is that they exert a con. stant.78 kn tangential force on the car in the direction of its motion, what is the magnitude of the car s velocity at position? From the solution to Problem 5.44, the work done by gravity is U gravity = ( Nm ) due to the change in elevation of W the car of h = m, and v g = Nm. The length of road between positions and is ( s = (3 π ) ( π ) 36.6 ) (3 ) 8 = 35. m. The work done by the tangential force is s U tgt = 78 ds = 78 ( 35. ) = N m. From the principle of work and energy U gravity + U tgt = from which ( ) W v g ( ) W v g, ( 9. 8 )( ) v = 4 = 6. m/s = 57.9 km/ h c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 99
22 Problem 5.46 The mass of the rocket is 5 kg. Its engine has a constant thrust of 45 kn. The length of the launching ramp is m. If the magnitude of the rocket s velocity when it reaches the end of the ramp is 5 m/s, how much work is done on the rocket by friction and aerodynamic drag? m U = U Fr+Dr + (45 kn)( m) (5 kg)(9.8 m/s )( m) U = (5 kg)(5 m/s) U Fr+Dr = 7 knm Problem 5.47 A bioengineer interested in energy requirements of sports determines from videotape that when the athlete begins his motion to throw the 7.5kg shot (Fig. a), the shot is stationary and.5 m above the ground. At the instant the athlete releases it (Fig. b), the shot is. m above the ground. The shot reaches a maximum height of 4.6 m above the ground and travels a horizontal distance of 8.66 m from the point where it was released. How much work does the athlete do on the shot from the beginning of his motion to the instant he releases it? Let v x and v y be the velocity components at the instant of release. Using the chain rule, a y = dv y dt = dv y dy dy dt = dv y dy v y = g, y and integrating, 4.6 v y dv y = g dy. v y.. m vy vx 4.6 m 8.66 m x v y = g(4.6.), wefind that v y = 7. m/s. The shot s x and y coordinates are given by x = v x t, y =. + v y t gt. Solving the first equation for t and substituting it into the second, ( ) x y =. + v y ( ) x v x g v x Setting x = 8.66 m, y = in this equation and solving for v x gives v x =. m/s. The magnitude of the shot s velocity at release is U = v x + v y = 3. m/s. Let U A be the work he does U A mg(y y ) = mv mv U A (7.5 kg)(9.8 m/s )(..5) = (7.5)(3.), or U A = 666 Nm c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
23 Problem 5.48 A small pellet of mass m =. kg starts from rest at position and slides down the smooth surface of the cylinder to position, where θ = 3. What work is done on the pellet as it slides from position to position? What is the magnitude of the pellet s velocity at position?.8 m u m v = R =.8 m U = mv mv R R cos 3 3 U = (.5)(.)v =.v The work is U = (.)(9.8)(.8.8 cos 3 ) =. Nm. U =. (Nm).v =. (Nm) v =.45 m/s Problem 5.49 In Active Example 5.4, suppose that you want to increase the value of the spring constant k so that the velocity of the hammer just before it strikes the workpiece is 4 m/s. what is the required value of k? Hammer The 4kg hammer is released from rest in position. The springs are unstretched when in position. Neglect friction. k k 4 mm U = mgh + kd = mv k = m d ( v gh k = ) ( 4 kg [4 m] ) (. m) [9.8m/s ][.4m] Workpiece 3 mm k = 48 N/m. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
24 Problem 5.5 Suppose that you want to design a bumper that will bring a 5 N. package moving at m/s to rest 5.4 mm from the point of contact with bumper. If friction is negligible, what is the necessary spring constant k? m/s k From the principle of work and energy, the work done on the spring must equal the change in kinetic energy of the package within the distance 5.4 m. ks = from which ( ) W v g k = ( W g ) ( v ) ( )( 5 = S ) = 6 N/m Problem 5.5 In Problem 5.5, what spring constant is necessary if the coefficient of kinetic friction between the package and the floor is µ k =.3 and the package contacts the bumper moving at m/s? The work done on the spring over the stopping dis tance is S S U S = Fds= ks ds = ks. The work done by friction over the stopping distance is S S U f = Fds= µ k Wds= µ k WS. From the principle of work and energy the work done must equal the kinetic energy of the package: ks + µ k WS = ( ) W v, g from which, for S =.5 m, k = ( ) W (v gµ k S) g S = 863 N/m c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
25 Problem 5.5 The 5N package starts from rest, slides down the smooth ramp, and is stopped by the spring. If you want the package to be brought to rest.5 m from the point of contact, what is the necessary spring constant k? What maximum deceleration is the package subjected to? 3 4 m k Find the spring constant N U = mgh kx = k = mgh x = (5 N)(4.5 m)sin 3 (.5m) k = 9 N/m The maximum deceleration occurs when the spring reaches the maximum compression (the force is then the largest). kx mg sin θ = ma a = k m x g sin θ (9 N/m) a = ( ) (.5 m) ( 9.8m/s ) sin 3 5 N 9. 8 m/s a = 83.4 m/s Problem 5.53 The 5N package starts from rest, slides down the smooth ramp, and is stopped by the spring. The coefficient of static friction between the package and the ramp is µ k =.. If you want the package to be brought to rest.5 m from the point of contact, what is the necessary spring constant k? 4 m k 3 Find the spring constant U = mgd sin θ µ k mg cos θ d kx = k = mgd x (sin θ µ k cos θ) (5 N)(4.5m) k = (.5 ) (sin 3. cos 3 ) m k = 73 N/ m. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 3
26 Problem 5.54 The system is released from rest with the spring unstretched. The spring constant k = N/m. Determine the magnitude of the velocity of the masses when the right mass has fallen m. 4 kg kg k When the larger mass falls m, the smaller mass rises m and the spring stretches m. For the system of two masses, springs, and the cable, U = ( ks)ds + ( m g) ds + m gds U = ks m gs + m gs U = k 4(9.8) + ()(9.8) U = Nm Also U = (m + m )V f m m 4 kg kg k Solving V f =.8 m/s Problem 5.55 The system is released from rest with the spring unstretched. The spring constant k = N/m. What maximum downward velocity does the right mass attain as it falls? From the solution to Problem 5.54, U = Ks + (m m )gs and U = (m + m )V For all s. Setting these equal, we get (m + m )V = (m m )gs Ks () Solve for dv ds and set dv ds to zero (m + m )v dv ds = (m m )g Ks = The extreme value for V occurs at S = (m m )g K =.785 m Substituting this back into () and solving, we get V =.7 m/s 4 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
27 Problem 5.56 The system is released from rest. The 4kg mass slides on the smooth horizontal surface. The spring constant is k = N/m, and the tension in the spring when the system is released is 5 N. By using the principle of work and energy, determine the magnitude of the velocity of the masses when the kg mass has fallen m. k 4 kg kg 5 N = ( N/m)x x =.5 m U = ( kg)(9.8 m/s )( m) ( N/m)([.5 m] [.5 m] ) U = (4 kg)(v ) v =.83 m/s Problem 5.57 Solve Problem 5.56 if the coefficient of kinetic friction between the 4kg mass and the horizontal surface is µ k =.4. 5 N = ( N/m)x x =.5 m U = ( kg)(9.8 m/s )( m) ( N/m)([.5 m] [.5 m] ) (.4)(4 kg)(9.8 m/s )( m) U = (4 kg)(v ) v =.59 m/s c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 5
28 Problem 5.58 The 4 N crate is released from rest on the smooth inclined surface with the spring unstretched. The spring constant is k = 8 N/ m. How far down the inclined surface does the crate slide before it stops? What maximum velocity does the crate attain on its way down? k 3 At an arbitrary distance s down the slope we have: U = (4 N)(s sin 3 ) (8 N/m)s = ( 4 N ) v 9.8 m/s When it stops, we set v = and solve for s = 5m Solving for v, we have v =.49(s 4s ) dv ds =.49( 8s) = s =.5 m Using s =.5 m v max = 3.5 m/s Problem 5.59 Solve Problem 5.58 if the coefficient of kinetic friction between the 4kg mass and the horizontal surface is µ k =.. At an arbitrary distance s down the slope we have: U = (4 N)(s sin 3 ) (8 N/m)s (.)(4 N cos 3 )s = ( 4 N ) v 9.8 m/s When it stops, we set v = and solve for s = 3.7 m Solving for v, we have v =.49(3.7s 4s ) dv =.49(3.7 8s) = ds s =.63 m Using s =.63 m v max =.9 m/s 6 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
29 Problem 5.6 The 4kg collar starts from rest in position with the spring unstretched. The spring constant is k = N/m. How far does the collar fall relative to position? k V = V f = Let position be the location where the collar comes to rest ks U = Ks + mgs k Also U = mv f mv = Thus = mgs Ks s(mg Ks) = Solving, s =.785 m. mg s c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 7
30 Problem 5.6 In position on the smooth bar, the 4kg collar has a downward velocity of m/s and the spring is unstretched. The spring constant is k = N/m. What maximum downward velocity does the collar attain as it falls? The work is U = Ks + mgs Also, k U = mv mv where m = 4 kg and V = m/s Thus, mv mv = Ks + mgs () Finding dv, and setting it to zero, ds mv dv ds = Ks + mg = s = mg/k =.39 m Solving () for V we get V =. m/s Problem 5.6 The 4kg collar starts from rest in position on the smooth bar. The tension in the spring in position is N. The spring constant is k = N/m. How far does the collar fall relative to position? For this problem, we need a new reference for the spring. If the tension in the spring is N T = kδ and K = N/m δ =. m. (the initial stretch) In this case, we have U = Ks s+. + mgs s+... Also V = V f = U = K(s +.) = + K(.) + mg(s +.) mg(.) Solving, we get s =.385 m 8 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
31 Problem 5.63 The 4kg collar is released from rest at position on the smooth bar. If the spring constant is k = 6 kn/m and the spring is unstretched in position, what is the velocity of the collar when it has fallen to position? Denote d = mm, h = 5 mm. The stretch of the spring in position is S = h + d d =. m and at S =. The work done by the spring on the collar is 5 mm k [ U spring = ( ks)ds = ]. ks = 43.3 Nm.. The work done by gravity is h U gravity = ( mg) ds = mgh = 9.8 Nm. From the principle of work and energy U spring + U gravity = mv, from which v = ( ) (U spring + U gravity ) = 5.5 m/s m mm Problem 5.64 The 4kg collar is released from rest in position on the smooth bar. The spring constant is k = 4 kn/m. The tension in the spring in position is 5 N. What is the velocity of the collar when it has fallen to position? Denote d = mm, h = 5 mm. The stretch of the spring at position is S = T k = 5 =.5 m. 4 The unstretched length of the spring is L = d S =..5 =.75 m. The stretch of the spring at position is S = h + d L =.45 m. The work done by the spring is U spring = S S ( ks)ds = k(s S ) = Nm. The work done by gravity is U gravity = mgh = 9.8 Nm. From the principle of work and energy is U spring + U gravity = mv, from which (Uspring + U gravity ) v = = 7.3 m/s m c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 9
32 Problem 5.65 The 4kg collar starts from rest in position on the smooth bar. Its velocity when it has fallen to position is 4 m/s. The spring is unstretched when the collar is in position. What is the spring constant k? The kinetic energy at position is mv = 3 Nm. From the solution to Problem 5.63, the stretch of the spring in position is S = h + d d =. m. The potential of the spring is U spring = ( ks)ds = S ks. The work done by gravity is U gravity = mgh = 9.8 Nm. From the principle of work of work and energy, U spring + U gravity = mv. Substitute and solve: k = ( ) mv U gravity S = 38 N/m Problem 5.66 The kg collar starts from rest at position and slides along the smooth bar. The yaxis points upward. The spring constant is k = N/m and the unstretched length of the spring is m. What is the velocity of the collar when it reaches position? y (4, 4, ) m (6,, ) m The stretch of the spring at position is (,, ) m S = (6 ) + ( ) + ( ) = 3. m. x The stretch of the spring at position is z S = (6 4) + ( 4) + ( ) = m. The work done by the spring is U spring = S S ( ks)ds = k(s S ) = 46.8 Nm. The work done by gravity is h U gravity = ( mg) ds = mgh = ()(9.8)(4 ) = 94.3 Nm. From the principle of work and energy: U spring + U gravity = mv, from which (Uspring + U gravity ) v = = 5.77 m/s m c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
33 Problem 5.67 A springpowered mortar is used to launch 44.5 N packages of fireworks into the air. The package starts from rest with the spring compressed to a length of 5. 4 mm. The unstretched length of the spring is 76 mm. If the spring constant is k =76 N/m, what is the magnitude of the velocity of the package as it leaves the mortar? 76 mm Equating the work done to the change in the kinetic energy, k(s S ) mg(y y ) = mv mv : 5.4 mm 6 ( 76 )[ (.6m ) ] ( 44.5)(.6sin 6 ft) = ( 44.5 / 9.8 )v. Solving, we obtain v = 39.3 m/s. Problem 5.68 Suppose that you want to design the mortar in Problem 5.67 to throw the package to a height of 45.7 m above its initial position. Neglecting friction and drag, determine the necessary spring constant. See the solution of Problem Let v be the velocity as the package leaves the barrel. To reach 45.7 m, mg( sin 6 ) = m(v sin 6 ). Solving, we obtain v = 34.4 m/s. Work and energy inside the barrel is k[ ( ).6m ] ( 44.5)(.6sin 6 m)= ( 44.5 / 9.8 )( 34.4 ), which gives k = 4549 N/m. Problem 5.69 Suppose an object has a string or cable with constant tension T attached as shown. The force exerted on the object can be expressed in terms of polar coordinates as F = T e r. Show that the work done on the object as it moves along an arbitrary plane path from a radial position r to a radial position r is U = T(r r ). r θ The work done on the object is U = r r F ds. T Suppose that the arbitrary path is defined by dr = (dre r + rdθe θ ), and the work done is r r U = F dr = T(e r e r )dr + r r r = Tdr= T(r r ) r since e r e s = bydefinition. r r T(re r e θ )rdθ c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
34 Problem 5.7 The kg collar is initially at rest at position. A constant N force is applied to the string, causing the collar to slide up the smooth vertical bar. What is the velocity of the collar when it reaches position? (See Problem 5.69.) mm The constant force on the end of the string acts through a distance s = =.3385 m. The work done by the constant force is U F = Fs = Nm. The work done by gravity on the collar is 5 mm N h U gravity = ( mg) ds = mgh = ()(9.8)(.5) = 9.8 Nm. From the principle of work and energy: U F + U gravity = mv, (UF + U gravity ) from which v = m = 4.9 m/s Problem 5.7 The kg collar starts from rest at position. The tension in the string is N, and the y axis points upward. If friction is negligible, what is the magnitude of the velocity of the collar when it reaches position? (See Problem 5.69.) y (,, ) m (4, 4, ) m N (6,, ) m The constant force moves a distance x s = (6 ) + ( ) + ( ) z (6 4) + ( 4) + ( ) =. m. The work done by the constant force is s U F = Fds= Fs = 439. Nm. The work done by gravity is h U gravity = ( mg) ds = mgh = ()(9.8)(3) = 94.3 Nm. From the principle of work and energy U F + U gravity = mv, from which (UF + U gravity ) v = = 5.38 m/s c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
35 Problem 5.7 As the F/A  8 lands at 64 m/s, the cable from A to B engages the airplane s arresting hook at C. The arresting mechanism maintains the tension in the cable at a constant value, bringing the 5.6 kn airplane to rest at a distance of m. What is the tension in the cable? (See Problem 5.69.) m U = T( ( m) + (. m). m) = ( ) 56 N ( [64 m/s] ) 9.8 m/s T = kn A C. m B Problem 5.73 If the airplane in Problem 5.7 lands at 73. m/s,what distance does it roll before the arresting system brings it to rest? U = ( 8585)( s + (. m). m) = ( ) 56 ( [64 m/s] ) 9.8 m/s Solving we find s = 6.6 m Problem 5.74 A spacecraft 3 km above the surface of the earth is moving at escape velocity v esc =,9 m/s. What is its distance from the center of the earth when its velocity is 5 percent of its initial value? The radius of the earth is 637 km. (See Example 5.6.) ( U = mgre ) = r r mv mv [ ] v r = v gre + r 3 km v esc [ (545 m/s) (,9 m/s) ] r = (9.8 m/s )(6,37, m) + 6,69, m r = 6,6 km. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 3
36 Problem 5.75 A piece of ejecta is thrown up by the impact of a meteor on the moon. When it is km above the moon s surface, the magnitude of its velocity (relative to a nonrotating reference frame with its origin at the center of the moon) is m/s. What is the magnitude of its velocity just before it strikes the moon s surface? The acceleration due to gravity at the surface of the moon is.6 m/s. The moon s radius is 738 km. km m/s The kinetic energy at h = km is [ m v] RM +h = 4 Nm. The work done on the ejecta as it falls from km is RM ( ) RM RM U ejecta = ( W ejecta )ds = mg M R M+h R M +h s ds = [ mg M R M s ] RM R M +h = mg M R M h R M + h, U ejecta =.8 m 6 Nm. From the principle of work and energy, at the Moon s surface: [ m [ m U ejecta = v] surface v] RM +h from which v surface = ( ) = 448 m/s 4 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
37 Problem 5.76 A satellite in a circular orbit of radius r around the earth has velocity v = gre /r, where R E = 637 km is the radius of the earth. Suppose you are designing a rocket to transfer a 9kg communication satellite from a circular parking orbit with 67km radius to a circular geosynchronous orbit with 4,km radius. How much work must the rocket do on the satellite? Denote the work to be done by the rocket by U rocket. Denote R park = 67 km, R geo = 4 km. The work done by the satellite s weight as it moves from the parking orbit to the geosynchronous orbit is Rgeo U transfer = Fds= R park = [ mg R E s ] Rgeo Rgeo R park ( R park = mgr E mg R E s ds ) ( ), R geo R park U transfer = Nm. From the principle of work and energy: [ ] [ ] U transfer + U rocket = mv geo mv. park from which [ ] [ ] U rocket = mv geo mv U transfer. park Noting [ ] mv = m geo ( gr E R geo ) = Nm, [ ] ( ) mv = m g R E =.67 Nm, park R park from which U rocket =.5 Nm Problem 5.77 The force exerted on a charged particle by a magnetic field is F = qv B, where q and v are the charge and velocity of the particle and B is the magnetic field vector. Suppose that other forces on the particle are negligible. Use the principle of work and energy to show that the magnitude of the particle s velocity is constant. The force vector F is given by a cross product involving v. This means that the force vector is ALWAYS perpendicular to the velocity vector. Hence, the force field does no work on the charged particle  it only changes the direction of its motion. Hence, if work is zero, the change in kinetic energy is also zero and the velocity of the charged particle is constant. c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 5
38 Problem 5.78 The N box is released from rest at position and slides down the smooth inclined surface to position. (c) If the datum is placed at the level of the floor as shown, what is the sum of the kinetic and potential energies of the box when it is in position? What is the sum of the kinetic and potential energies of the box when it is in position? Use conservation of energy to determine the magnitude of the box s velocity when it is in position. 3 m M5 m Datum T + V = + ( N)(5 m) = 5 Nm T + V = T + V = 5 Nm (c) 5  = ( ) N N m v + ( N)( m) v = 7.67 m/s 9.8 m/s Problem 5.79 The.45kg soccer ball is m above the ground when it is kicked upward at m/s. Use conservation of energy to determine the magnitude of the ball s velocity when it is 4 m above the ground. Obtain the answer by placing the datum at the level of the ball s initial position and at ground level. m/s m Datum m/s m Datum T = (.45 kg)( m/s), V = T = (.45 kg)v,v = (.45 kg)(9.8 m/s )(3 m) T + V = T + V v = 9.3 m/s T = (.45 kg)( m/s), V = (.45 kg)(9.8 m/s )( m) T = (.45 kg)v,v = (.45 kg)(9.8 m/s )(4 m) T + V = T + V v = 9.3 m/s 6 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
39 Problem 5.8 The Lunar Module (LM) used in the Apollo moon landings could make a safe landing if the magnitude of its vertical velocity at impact was no greater than 5 m/s. Use conservation of energy to determine the maximum height h at which the pilot could shut off the engine if the vertical velocity of the lander is m/s downward and m/s upward. The acceleration due to gravity at the moon s surface is.6 m/s. h Use conservation of energy Let state be at the max height and state at the surface. Datum is at the lunar surface., mv + mgh = mv + v = 5 m/s g =.6 m/s v =± m/s (m cancels from the equation.) h = g (v v ) (The sign of V does not matter since v is the only occurrence of v in the relationship). Solving h = 6.48 m Problem 5.8 The.4kg collar starts from rest at position and slides down the smooth rigid wire. The y axis points upward. Use conservation of energy to determine the magnitude of the velocity of the collar when it reaches point. y (5, 5, ) m.4 kg x z (3,,) m Assume gravity acts in the y direction and that y = is the datum. By conservation of energy, mv + V = constant where V = mgy. Thus, mv + mgy = mv + mgy m =.4 kg, g = 9.8 m/s, v =, y =, and y = 5 m. Thus y (5, 5, ) m.4 kg + (.4)(9.8)(5) = (.4)v + v = 9.9 m/s x z (3,, ) m c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior 7
40 Problem 5.8 At the instant shown, the kg mass is moving downward at.6 m/s. Let d be the downward displacement of the mass relative to its present position. Use conservation of energy to determine the magnitude of the velocity of the kg mass when d = m. m = 4kg m = kg 4 kg kg v =.6 m/s V g = 9.8 m/s d = m.6 m s.6 m/s A Energy for the system is conserved 4 kg A B kg d V ( m v + ) + ( m v + ) = m v + m v + m g(d) m g(d) (m + m )v = (m + m )v + (m m )gd Datum State () V A = V B = B State () d Substituting known values and solving v = 3.95 m/s Problem 5.83 The mass of the ball is m = kg and the string s length is L = m. The ball is released from rest in position and swings to position, where θ = 4. Use conservation of energy to determine the magnitude of the ball s velocity at position. Draw graphs of the kinetic energy, the potential energy, and the total energy for values of θ from zero to 8. m u L m = kg L = m Use conservation of energy State θ = ; State, <θ<8 Datum: θ =, v =, g = 9.8 m/s mv + mg() = mv + mg( L sin θ) KE = mv V = mgl sin θ for all θ. Total energy is always zero (datum value). KE, PE, and (PE + KE) Nm Kinetic and Potential Energy vs α KE positive, PE negative, (KE + PE) zero 5 KE PE 5 TOT Theta (degrees) Evaluating at θ = 4, v = 3.55 m/s 8 c 8 Pearson Education South Asia Pte Ltd. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior
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