Solution to HW#6 Phys 211 Fall 2002

Transcription

1 Solution to H#6 Phys Fall 8. RASONING AND SOLUTION The speed o a particle doubles and then doubles again, because a net external orce acts on it. Let m and v 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 = 3 mv K K During the second doubling, the change in the kinetic energy o the particle is = m v m v = mv K K 4 6 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. 5. RASONING AND SOLUTION Car A turns o its engine and coasts up the hill. Car B keeps its engine running and drives up the hill at constant speed. I air resistance and riction are negligible, then only the motion o car A is an example o the principle o conservation o mechanical energy. During the motion o car A, there are no nonconservative orces present. As car A coasts up the hill it loses kinetic energy and gains an equal amount o gravitational potential energy. The motion o car B is not described by the principle o conservation o mechanical energy, because the driving orce that propels car B up the hill is a nonconservative orce. 6.P.6. RASONING AND SOLUTION a. In both cases, the lit orce L is perpendicular to the displacement o the plane, and, thereore, does no work. As shown in the drawing in the text, when the plane is in the dive, there is a component o the weight that points in the direction o the displacement o the plane. hen the plane is climbing, there is a component o the weight that points opposite to the displacement o the plane. Thus, since the thrust T is the same or both cases, the net orce in the direction o the displacement is greater or the case where the plane is diving. Since the displacement s is the same in both cases, more net work is done during the dive. b. The work done during the dive is dive = (T + cos 75 ) s, while the work done during the climb is climb = (T + cos 5 ) s. Thereore, the dierence between the net work done during the dive and the climb is

2 dive climb = (T + cos 75 ) s (T + cos 5 ) s = s (cos 75 cos 5 ) = (5.9 4 N)(.7 3 m)(cos 75 cos 5 ) = J 6.P.9. SSM RASONING AND SOLUTION According to quation 6., the work done by the husband and wie are, respectively, [ ] H FH θh Husband = ( cos ) s [ ] F θ ie = ( cos ) s Since both the husband and the wie do the same amount o work, ( F cos θ ) s = ( F cos θ ) s H H Since the displacement has the same magnitude s in both cases, the magnitude o the orce exerted by the wie is F cosθ cos 58 (67 N) 45 N H = FH = = cosθ cos 38 6.P.6. RASONING AND SOLUTION From the work-energy theorem, quation 6.3, = mv mv = m v v a. (74 kg) (845 m/s) (8 m/s) = =.35 J b. (74 kg) (8 m/s) (845 m/s) = =.35 J 6.P.. RASONING It is useul to divide this problem into two parts. The irst part involves the skier moving on the snow. e can use the work-energy theorem to ind her speed when she comes to the edge o the cli. In the second part she leaves the snow and alls reely toward the ground. e can again employ the work-energy theorem to ind her speed just beore she lands. SOLUTION The drawing at the right shows the three orces that act on the skier as she glides on the snow. The orces are: her weight mg, the normal orce F N, and the kinetic rictional orce k. Her displacement is labeled as s. The work-energy theorem, quation 6.3, is k 65. F N s +y 5. mg

3 = mv mv where is the work done by the net external orce that acts on the skier. The work done by each orce is given by quation 6., = F cos θ s, so the work-energy theorem becomes ( mg cos 65. ) s + ( cos 8 ) s + ( F cos 9 ) s = mv mv k N Since cos 9 =, the third term on the let side can be eliminated. The magnitude k o the kinetic rictional orce is given by quation 4.8 as k = µ kfn. The magnitude F N o the normal orce can be determined by noting that the skier does not leave the surace o the slope, so a y = m/s. Thus, we have that ΣF y =, so F mg cos 5. = or F = mg cos 5. N ΣF y N The magnitude o the kinetic rictional orce becomes k = µ kfn = µ kmg cos 5.. Substituting this result into the work-energy theorem, we ind that ( mg cos 65. ) s + ( µ mg cos 5. )( cos 8 ) s = mv mv k Algebraically eliminating the mass m o the skier rom every term, setting cos 8 = and v = m/s, and solving or the inal speed v, gives v = gs cos 65. µ cos 5. k = 9.8 m/s.4 m cos 65.. cos 5. = 7. m/s The drawing at the right shows her displacement s during ree all. Note that the displacement is a vector that starts where she leaves the slope and ends where she touches the ground. The only orce acting on her during the ree all is her weight mg. The workenergy theorem, quation 6.3, is θ 3.5 m s mg = mv mv The work is that done by her weight, so the work-energy theorem becomes

4 ( mg cosθ ) s = mv mv In this expression θ is the angle between her weight (which points vertically downward) and her displacement. Note rom the drawing that s cos θ = 3.5 m. Algebraically eliminating the mass m o the skier rom every term in the equation above and solving or the inal speed v gives v = v + g scosθ = 7. m/s m/s 3.5 m =.9 m/s 6.P.8. RASONING AND SOLUTION a. From the deinition o work, = Fs cos θ. For upward motion θ = 8 so = mgs = (7. N)(.3 m.5 m) = 43 J b. The change in potential energy is P = mgh mgh = (7. N)(.3 m.5 m) = +43 J 6.P.36. RASONING Since air resistance is being neglected, the only orce that acts on the gol ball is the conservative gravitational orce (its weight). Since the maximum height o the trajectory and the initial speed o the ball are known, the conservation o mechanical energy can be used to ind the kinetic energy o the ball at the top o the highest point. The conservation o mechanical energy can also be used to ind the speed o the ball when it is 8. m below its highest point. SOLUTION a. The conservation o mechanical energy, quation 6.9b, states that mv + mgh = mv + mgh K Solving this equation or the inal kinetic energy, K, yields K = mv + mg h h =.47 kg 5. m / s +.47 kg 9.8 m / s m 4.6 m = 5. J b. The conservation o mechanical energy, quation 6.9b, states that

5 mv + mgh = mv + mgh The mass m can be eliminated algebraically rom this equation, since it appears as a actor in every term. Solving or v, and noting that the inal height o the ball is h = 4.6 m 8. m = 6.6 m, we have that v = v + g h h = 5. m / s m / s m 6.6 m = 48.8 m / s 6.P.4. SSM RASONING Friction and air resistance are being ignored. The normal orce rom the slide is perpendicular to the motion, so it does no work. Thus, no net work is done by nonconservative orces, and the principle o conservation o mechanical energy applies. SOLUTION Applying the principle o conservation o mechanical energy to the swimmer at the top and the bottom o the slide, we have mv + mgh = mv + mgh I we let h be the height o the bottom o the slide above the water, h = h, and h = H. Since the swimmer starts rom rest, v = m/s, and the above expression becomes Solving or H, we obtain v + gh = gh H = h + v g Beore we can calculate H, we must ind v and h. Since the velocity in the horizontal direction is constant, v = x t = 5. m =. m/s.5 s The vertical displacement o the swimmer ater leaving the slide is, rom quation 3.5b (with down being negative), y = a yt = ( 9.8 m/s )(.5 s) =.3 m

6 Thereore, h =.3 m. Using these values o v and h in the above expression or H, we ind H = h + v (. m/s) =.3 m + g (9.8 m/s ) = 6.33 m 6.P.45. RASONING AND SOLUTION The work-energy theorem can be used to determine the change in kinetic energy o the car according to quation 6.8: ( ) ( ) nc = mv + mgh mv + mgh The nonconservative orces are the orces o riction and the orce due to the chain mechanism. Thus, we can rewrite quation 6.8 as ( + = K + mgh K + mgh riction chain e will measure the heights rom ground level. Solving or the change in kinetic energy o the car, we have K K = + mgh + mgh riction chain ) 4 4 =. J + 3. J (375 kg)(9.8 m/s )(. m) + ( kg)(9.8 m/s )(5. m) = 4.5 J 6.P.49. RASONING AND SOLUTION The orce exerted by the bat on the ball is the only non-conservative orce acting. The work due to this orce is nc = + mv mv mg h h Taking h = m at the level o the bat, v = 4. m/s just beore the bat strikes the ball and v to be the speed o the ball at h = 5. m, we have v v gh m nc = + (7. J) + 4. m/s (9.8 m/s )(5. m) 45.9 m/s = =.4 kg