grav mgr, where r is the radius of the bowl and grav W mgr kg 9.8 m s m J.

Transcription

1 Phys 0 Homework 9 Solutions 3. (a) The force of ity is constant, so the work it does is given by W F d, where F is the force and d is the displacement. The force is vertically downward and has magnitude mg, where m is the mass of the flake, so this reduces to the height from which the flake falls. Thus W mgr, where r is the radius of the bowl and 3 3 W mgr.00 0 kg 9.8 m s 0.0 m J. (b) The force of ity is conservative, so the change in itational potential energy of the flake-earth system is the negative of the work done: 3 W J. (c) If 0 J at the bottom, then at the top the potential energy is J. (d) If 0 J at the top, then at the bottom the potential energy is J. (e) All the answers are proportional to the mass of the flake. If the mass is doubled, all answers are doubled. 4. In each case, we use the fact that W mg y. (a) To point A, the work done by ity is W mg y 0 J.

2 (b) To point B, the work done by ity is W mg y mg h mgh. (c) To point C, the work done by ity is W mg y mg h mgh. (d) With C 0 J., we obtain mgh B. (e) With C 0 J., we obtain A mgh. (f) All the answers are proportional to the mass of the object. If the mass is doubled, all answers are doubled. 5. (a) The only force that does work on the ball is the force of ity; the force of the rod is perpendicular to the path of the ball and so does no work. In going from its initial position to the lowest point on its path, the ball moves vertically through a distance equal to the length L of the rod, so the work done by the force of ity is W mgl. (b) In going from its initial position to the highest point on its path, the ball moves vertically through a distance equal to L, but this time the displacement is upward, opposite the direction of the force of ity. The work done by the force of ity is W mgl. (c) The final position of the ball is at the same height as its initial position. The displacement is horizontal, perpendicular to the force of ity. The force of ity does no work during this displacement. (d) The force of ity is conservative. The change in the itational potential energy of the ball-earth system is the negative of the work done by ity: the lowest point. mgl as the ball goes to

3 (e) Similarly, we find mgl as it goes to the highest point. (f) Furthermore, we have 0 J as it goes to the point at the same height. (g) The change in the itational potential energy depends only on the initial and final positions of the ball, not on its speed anywhere. The change in the potential energy is the same since the initial and final positions are the same. 7. (a) The force of ity is constant, so the work it does is given by W F d, where F is the force and d is the displacement. The force is vertically downward and has magnitude mg, where m is the mass of the snowball. The epression for the work reduces to the height through which the snowball drops. Thus W mgh, where h is W mgh.50 kg 9.8 m s.5 m 84 J. (b) The force of ity is conservative, so the change in the potential energy of the snowball- Earth system is the negative of the work it does: W 84 J. (c) If the potential energy on the cliff is taken to be 0 J, the potential energy on the ground is 84 J. 5. The initial speed of the truck is 5 with the horizontal. 3 v 30 0 m 3600 s 36 m s. The ramp makes an angle (a) The change in the truck s potential energy as it climbs the ramp is conserve energy: K K, which leads to truck to come to a stop, so v 0 m s. Therefore, sin. mgl sin. We m v v mgl We want the

4 L v 36 m s g sin 9.8 m s sin5 60 m. (b) The required length does not depend on the mass of the truck. It remains the same if the mass is reduced. (c) If the speed is decreased, L decreases since L v. 7. (a) The change in potential energy as the snowball falls a distance h.5 m is ground is mgh. From conservation of mechanical energy, the kinetic energy when it hits the K K mv mgh. Therefore the speed of the snowball just as it hits the ground is 4.0 m s 9.8 m s.5 m.0 m s. v v gh (b) The speed does not depend on the direction of the initial velocity, only the magnitude of the initial velocity. There would be no change in the speed. (c) The speed does not depend on the mass of the snowball. Thus, changing the mass of the snowball does not change the speed. 9. (a) The change in the itational potential energy is 3 mg y marble kg 9.8 m s 0 m 0.98 J.

5 (b) Since the kinetic energy is zero at the release point and at the highest point, then conservation of mechanical energy requires elas elas 0 J. Therefore, 0.98 J. (c) Taking the potential energy of the relaed spring to be zero, the compressed spring then has a elas elas potential energy of 0.98 J. The spring constant is therefore k y elas spring 0.98 J m 30 N m.. Set the positive ais pointing up the ramp. (a) Setting the zero of elastic potential energy as the relaed spring, the elastic potential energy of the compressed spring is k spring 960 N m 0.00 m 39. J. (b) At the highest point on the incline, the kinetic energy and the spring s elastic potential energy will be zero. Since the block starts at rest, the change in itational potential energy is spring spring 39. J. (c) The distance traveled up the ramp is. Since the ramp is at an angle of 30 to the horizontal, the vertical component of that displacement is h sin. The change in itational potential energy is along the ramp is mgh mg sin 39. J. Therefore, the distance mg 39. J sin.00 kg 9.8 m s sin m.

6 3. Elastic potential energy of the spring is zero when the spring is relaed. The spring constant of the spring is k 70 N 0.00 m 3500 N m. The initial position of the block will be, the position where the block just touches the spring will be 0 m, and the stopping point will be. 3 We ll set the positive direction to be down the ramp. (a) At the end of this process, the spring is compressed by cm. The potential energy of spring the spring is now k 3500 N m m 0.4 J. The block started from rest 3 3 and stops at the end, so the kinetic energy doesn t change. Therefore, the itational potential energy must change by spring 0.4 J. That change in itational potential energy occurs because the block ends its motion a vertical distance h sin30 below 3 where it started. Since mgh, the block moves down the incline by a distance of 0.4 J mg sin 30 kg 9.8 m s sin m. (b) When the block just touches the spring, it has traveled a distance m m 0.9 m. The itational potential energy of the block has changed by mg sin30. Conserving mechanical energy leads to K mv mg sin30.

7 Therefore, v g sin m s 0.9 m sin30.7 m s. 5. We denote m as the mass of the block, h = 0.40 m as the height from which it dropped (measured from the relaed position of the spring), and the compression of the spring (measured downward so that it yields a positive value). The kinetic energy is zero both at the beginning and end of this motion. Conservation of mechanical energy leads to mg h k spring. Therefore, the distance may be found from k mg mgh Noting that 0. mg.0 kg 9.8 m s 9.6 N, we use the quadratic formula, choosing the positive root, to obtain mg mg mghk k 9.6 N 9.6 N 9.6 N 0.40 m 960 N m 960 N m 0.0 m. 7. With our origin at the launching point, let the positive ais point to the right, and the positive y ais up. When the spring is compressed by a distance d, the potential energy of the spring is kd where k is the spring constant. The launching process conserves spring, mechanical energy, so the kinetic energy of the marble is K mv kd so the initial, velocity of the marble is in the + direction and has a magnitude of v d k m. The - component of velocity is a constant in this motion, so the horizontal position on the ground is given by v t t k m d. Whether Bobby or Rhoda is playing this game, the time to impact is the same t y g. As a result, the horizontal position on the ground is proportional to the compression of the spring. Therefore, Rhoda should compress the spring by

8 d Rhoda d Bobby Rhoda Bobby.0 m.0 cm.5 cm..0 m 0.70 m