Homework #4
Conceptual Solution

1. (c). At point II, the child has gravitational potential energy (being at some height above the ground), as well as kinetic energy (being moving with some velocity down the slide). At point IV, the child is at ground level (no gravitational potential energy) and has stopped (no kinetic energy). Thus at point IV the child has no mechanical energy; it has all been converted to heat, by the frictional force actiing between III and IV.
2. (c). The designers must stop the child, coming down the same slide, using a frictional force over half the distance. To stop the child, the frictional force must do work draining all of the child's kinetic energy. But the child's mechanical energy is conserved on the slide (where only the conservative force of gravity acts), and so the child's kinetic energy at III is equal its gravitational potential energy at I. It thus depends only on the slide's height, and stays the same when engineers redesign the stopping surface. This same kinetic energy must be the negative work done by friction between III and IV, which reduces the child's kinetic energy at point IV to zero. Thus the work required by friction remains the same.
3. (a). From the previous questions, friction must do the same work W in half the distance d. Since W = Fd, we can cut d in half and keep W the same only by doubling the force F, to keep the product Fd the same.
4. (a). The power output of each child is the work done per unit time: P = W/t. The work done by each child in climbing the slide gives his change in mechanical energy -- here just the change in gravitational potential energy from bottom to top, mgh, which is the same for each (equal mass) child. But child A takes half as long to do this work as child B, thus

   

   This makes intuitive sense: doing the same work in less time requires more energy per unit time, since less time is available to do the work.

5. (b) As explained in question 4, both children gain the same potential energy; thus they do the same work.
6. (c). If we take air resistance into account, air resistance does nonconservative negative work on the child, reducing his mechanical energy, . But total energy is conserved; this work serves only to transfer energy from the child (in the form of mechanical energy) to the air (in the form of heat), and is thus equal and opposite the work done by the child on the air. The magnitude of this work is thus the child's change in mechanical energy, which is the change in kinetic energy PLUS the change in potential energy.
7. (a) Since the child travels too fast at point II, she leaves the circle. At that moment, she was traveling along the circle, in the tangential -- here horizontal -- direction. However, as she continues to travel, gravity acts on her, accelerating her downward. This combination of initial horizontal motion with downward acceleration is choice (a).
8. (d) We know that the magnitude of the work done by friction, , in stopping the child must equal the child's kinetic energy at point III. Since the slide is frictionless, that kinetic energy at point III, the bottom, must equal the potential energy mgh at I, the top, where the child is at rest, by conservation of mechanical energy. Thus

   

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