= Ps cos 0 = (150 N)(7.0 m) = J F N. s cos 180 = µ k

Transcription

1 Week 5 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions o these problems, various details have been changed, so that the answers will come out dierently. The method to ind the solution is the same, but you will need to repeat part o the calculation to ind out what your answer should have been. WebAssign Problem 1: During a tug-o-war, team A pulls on team B by applying a orce o 1100 N to the rope between them. How much work does team A do i they pull team B toward them a distance o 2.0 m? REASONING AND SOLUTION We will assume that the tug-o-war rope remains parallel to the ground, so that the orce that moves team B is in the same direction as the displacement. According to Equation 6.1, the work done by team A is W = (F cos )s = (1100 N)(cos 0 )(2.0 m) = J WebAssign Problem 2: A 55-kg box is being pushed a distance o 7.0 m across the loor by a orce whose magnitude is 150 N. The orce is parallel to the displacement o the box. The coeicient o kinetic riction is Determine the work done on the box by each o the our orces that act on the box. Be sure to include the proper plus or minus sign or the work done by each orce. REASONING AND SOLUTION The applied orce does work W P = Ps cos 0 = (150 N)(7.0 m) = J The rictional orce does work W = k s cos 180 = µ k F N s where F N = mg, so W = (0.25)(55 kg)(9.80 m/s 2 )(7.0 m) = 940 J, since they both act at a 90 angle to the displacement. WebAssign Problem 3: The concepts in this problem are similar to those in Multiple- Concept Example 4, except that the orce doing the work in this problem is the tension in the cable. A rescue helicopter lits a 79-kg person straight up by means o a cable. The person has an upward acceleration o 0.70 m/s 2 and is lited rom rest through a distance o 11 m. (a) What is the tension in the cable? How much work is done by (b) the tension

2 in the cable and (c) the person s weight? (d) Use the work energy theorem and ind the inal speed o the person. REASONING Since the person has an upward acceleration, there must be a net orce acting in the upward direction. The net orce ΣF y is related to the acceleration a y by Newton s second law, Σ Fy = may, where m is the mass o the person. This relation will allow us to determine the tension in the cable. The work done by the tension and the person s weight can be ound directly rom the deinition o work, Equation 6.1. SOLUTION a. The ree-body diagram at the right shows the two orces that act on the person. Applying Newton s second law, we have T s +y T mg = ma Σ F y y Solving or the magnitude o the tension in the cable yields mg T = m(a y + g) = (79 kg)(0.70 m/s m/s 2 ) = N b. The work done by the tension in the cable is 2 3 ( ) WT = T cos s = ( N) (cos 0 ) (11 m) = J (6.1) c. The work done by the person s weight is ( ) ( ) W = mg s = (6.1) 2 3 W cos (79 kg) 9.8 m/s (cos 180 ) (11 m) = J d. The work-energy theorem relates the work done by the two orces to the change in the kinetic energy o the person. The work done by the two orces is W = W T + W W : W T + WW = mv 2 mv (6.3) W Solving this equation or the inal speed o the person gives

3 2 v = v + W + W m ( ) 2 0 T W = ( 0 m/s ) + ( J J ) = 4 m / s 79 kg WebAssign Problem 4: A 0.60-kg basketball is dropped out o a window that is 6.1 m above the ground. The ball is caught by a person whose hands are 1.5 m above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy o the basketball, relative to the ground, when it is (b) released and (c) caught? (d) How is the change (PE PE 0 ) in the ball s gravitational potential energy related to the work done by its weight? REASONING The work done by the weight o the basketball is given by Equation 6.1 as W = ( F cos ) s, where F = mg is the magnitude o the weight, is the angle between the weight and the displacement, and s is the magnitude o the displacement. The drawing shows that the weight and displacement are parallel, so that = 0. The potential energy o the basketball is given by Equation 6.5 as PE = mgh, where h is the height o the ball above the ground. SOLUTION a. The work done by the weight o the basketball is ( cos ) W = F s = mg (cos 0 )(h 0 h ) = (0.60 kg)(9.80 m/s 2 )(6.1 m 1.5 m) = 27 J is b. The potential energy o the ball, relative to the ground, when it is released PE 0 = mgh 0 = (0.60 kg)(9.80 m/s 2 )(6.1 m) = 36 J (6.5) c. The potential energy o the ball, relative to the ground, when it is caught is PE = mgh = (0.60 kg)(9.80 m/s 2 )(1.5 m) = 8.8 J (6.5) d. The change in the ball s gravitational potential energy is PE = PE PE 0 = 8.8 J 36 J = 27 J We see that the change in the gravitational potential energy is equal to 27 J = W, where W is the work done by the weight o the ball (see part a).

4 WebAssign Problem 5: A person starts rom rest at the top o a large rictionless spherical surace, and slides into the water below (see the drawing). At what angle does the person leave the surace? (Hint: When the person leaves the surace, the normal orce is zero.) REASONING AND SOLUTION I air resistance is ignored, the only nonconservative orce that acts on the person is the normal orce exerted on the person by the surace. Since this orce is always perpendicular to the direction o the displacement, the work done by the normal orce is zero. We can conclude, thereore, that mechanical energy is conserved. (1) 1 2 mv mgh 0 = 1 2 mv 2 + mgh where the inal state pertains to the position where the person leaves the surace. Since the person starts rom rest v 0 = 0 m/s. Since the radius o the surace is r, h 0 = r, and h = r cos where is the angle at which the person leaves the surace. Equation (1) becomes r r r cos mgr = 1 2 mv 2 + mg(r cos ) (2) In general, as the person slides down the surace, the two orces that act on him are the normal orce F N and the weight mg. The centripetal orce required to keep the person moving in the circular path is the resultant o F N and the radial component o the weight, mg cos. When the person leaves the surace, the normal orce is zero, and the radial component o the weight provides the centripetal orce. mg cos mg F N mg cos = mv 2 r v 2 = gr cos (3)

5 Substituting this expression or v 2 into Equation (2) gives Solving or gives 1 2 mgr = mg( r cos ) + mg( r cos ) 1 2 = cos = 48 3 WebAssign Problem 6: Bicyclists in the Tour de France do enormous amounts o work during a race. For example, the average power per kilogram generated by Lance Armstrong (m = 75.0) is 6.50 W per kilogram o his body mass. (a) How much work does he do during a 135-km race in which his average speed is 12.0 m/s? (b) Oten, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms o nutritional Calories, noting that nutritional Calories. REASONING The work W done is equal to the average power P multiplied by the time t, or W = Pt (6.10a) The average power is the average power generated per kilogram o body mass multiplied by Armstrong s mass. The time o the race is the distance s traveled divided by the average speed v, or t = s / v (see Equation 2.1). SOLUTION a. Substituting t = s / v into Equation 6.10a gives s W = Pt = P = v ( ) W kg kg P m 12.0 m/s 6 = J b. Since 1 joule = nutritional calories, the work done is ( ) 6 6 W = joules = joules nutritional calories 1 joule 3 = nutritional calories

6 WebAssign Problem 7: Suppose in Figure 6.2 that o work are done by the orce ( ) in moving the suitcase a distance o 50.0 m. At what angle is the orce oriented with respect to the ground? REASONING AND SOLUTION Solving Equation 6.1 or the angle, we obtain 3 1 W J = cos = cos = 42.8 F s (30.0 N)(50.0 m) WebAssign Problem 8: The drawing shows a version o the loop-the-loop trick or a small car. I the car is given an initial speed o 4.0 m/s, what is the largest value that the radius r can have i the car is to remain in contact with the circular track at all times? REASONING AND SOLUTION When the car is at the top o the track the centripetal orce consists o the ull weight o the car. mv 2 /r = mg gives Applying the conservation o energy between the bottom and the top o the track (1/2)mv mg(2r) = (1/2)mv 0 Using both o the above equations 2 v 0 = 5gr so

7 r = v 0 2 /(5g) = (4.00 m/s) 2 /(49.0 m/s 2 ) = m Practice conceptual problems: 5. A ball has a speed o 15 m/s. Only one external orce acts on the ball. Ater this orce acts, the speed o the ball is 7 m/s. Has the orce done positive or negative work? Explain. REASONING AND SOLUTION The speed o the ball decreases; thereore, the ball is subjected to an external resistive orce. A resistive orce always points opposite to the direction o the displacement o the ball. Thereore, the external orce does negative work. Using the work-energy theorem, we see that the change in the kinetic energy o the ball is negative; thereore, the total work done on the ball is negative. We can conclude, thereore, that the net orce did negative work on the ball. 8. The speed o a particle doubles and then doubles again because a net external orce acts on it. Does the net orce do more work during the irst or the second doubling? Justiy your answer. REASONING AND SOLUTION The speed o a particle doubles and then doubles again, because a net external orce acts on it. Let m and v 0 represent the mass and initial speed o the particle, respectively. During the irst doubling, the change in the kinetic energy o the particle is ( ) = m v mv = mv KE KE 2 During the second doubling, the change in the kinetic energy o the particle is ( ) ( ) = m v m v = mv0 KE KE From the work-energy theorem, we know that a change in kinetic energy is equal to the work done by the net external orce. Thereore, more work is done by the net orce during the second doubling. 14. A person is riding on a Ferris wheel. When the wheel makes one complete turn, is the net work done by the gravitational orce positive, negative, or zero? Justiy your answer. REASONING AND SOLUTION As the person moves downward rom the top o the Ferris wheel, his displacement points downward. Since the person s weight also points downward, the work done by gravity is positive. As the person moves upward rom the bottom, his displacement points upward. Since the weight still points

8 downward, the work done by gravity is negative. The magnitude o the work done in each hal cycle is the same; thereore, the net work done in one revolution is zero. 17. The drawing shows an empty uel tank about to be released by three dierent jet planes. At the moment o release, each plane has the same speed and each tank is at the same height above the ground. However, the directions o travel are dierent. In the absence o air resistance, do the tanks have dierent speeds when they hit the ground? I so, which tank has the largest speed and which has the smallest speed? Explain. REASONING AND SOLUTION Since each plane has the same speed, the kinetic energy o each uel tank will be the same at the instant o release. Since each plane is at the same height above the ground, each uel tank must all through the same vertical displacement. Thereore, the work done by gravity on each uel tank is the same. From the work-energy theorem, each uel tank will gain the same amount o kinetic energy during the all. Thereore, each uel tank will hit the ground with the same speed.