Physics 16 Problem Set 7 Solutions

Physics  16  Problem  Set  7  Solutions   Y&F  Problems     7.37. IDENTIFY and SET UP: Use Eq.(7.17) to calculate the force from U. At equilibrium (a) EXECUTE: The graphs are sketched in Figure 7.37. Figure 7.37 (b) At equilibrium so implies solution is the equilibrium distance U is a minimum at this r; the equilibrium is stable. (c) At At The energy that must be added is (d) gives that and and Then EVALUATE: gives As the graphs in part (a) show, is the slope of at each r. has a minimum where   7.38. IDENTIFY: SET UP: EXECUTE: Apply Eq.(7.16). is the slope of the U versus x graph. (a) Considering only forces in the x-direction, and so the force is zero when the slope of the U vs x graph is zero, at points b and d. (b) Point b is at a potential minimum; to move it away from b would require an input of energy, so this point is stable. (c) Moving away from point d involves a decrease of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point. EVALUATE: At point b, is negative when the marble is displaced slightly to the right and is positive when the marble is displaced slightly to the left, the force is a restoring force, and the equilibrium is stable. At point d, a small displacement in either direction produces a force directed away from d and the equilibrium is unstable.   7.42. IDENTIFY: Apply Eq.(7.14). SET UP: Only the spring force and gravity do work, so . Let at the horizontal surface. EXECUTE: (a) Equating the potential energy stored in the spring to the block's kinetic energy, or (b) Using energy methods directly, the initial potential energy of the spring equals the final gravitational potential energy, or EVALUATE: The total energy of the system is constant. Initially it is all elastic potential energy stored in the spring, then it is all kinetic energy and finally it is all gravitational potential energy.   7.63. IDENTIFY and SET UP: First apply to the skier. Find the angle where the normal force becomes zero, in terms of the speed v2 at this point. Then apply the work-energy theorem to the motion of the skier to obtain another equation that relates v2 and Solve these two equations for Let point 2 be where the skier loses contact with the snowball, as sketched in Figure 7.63a Loses contact implies Figure 7.63a First, analyze the forces on the skier when she is at point 2. The free-body diagram is given in Figure 7.63b. For this use coordinates that are in the tangential and radial directions. The skier moves in an arc of a circle, so her acceleration is directed in towards the center of the snowball. EXECUTE: But so Figure 7.63b Now use conservation of energy to get another equation relating The only force that does work on the skier is gravity, so Then Combine this with the so equation: and to EVALUATE: She speeds up and her increases as she loses gravitational potential energy. She loses contact when she is going so fast that the radially inward component of her weight isn’t large enough to keep her in the circular path. Note that where she loses contact does not depend on her mass or on the radius of the snowball.   7.67. and (a) IDENTIFY: function Use Eq.(6.7) to calculate W and then use to identify the potential energy SET UP: Let and Let be some arbitrary point x, so EXECUTE: EVALUATE: If the spring does obey Hooke’s law, with (b) IDENTIFY: Apply Eq.(7.15) to the motion of the object. SET UP: The system at points 1 and 2 is sketched in Figure 7.67. and our result reduces to The only force that does work on the object is the spring force, so Figure 7.67 EXECUTE: Thus EVALUATE: The elastic potential energy stored in the spring decreases and the kinetic energy of the object increases.   7.75. (a) IDENTIFY and SET UP: Apply to the motion from A to B. EXECUTE: where Thus (The work done by F goes partly to the potential energy of the stretched spring and partly to the kinetic energy of the block.) and Thus and (b) IDENTIFY: Apply Eq.(7.15) to the motion of the block. Let point C be where the block is closest to the wall. When the block is at point C the spring is compressed an amount so the block is from the wall, and the distance between B and C is SET UP: The motion from A to B to C is described in Figure 7.75. EXECUTE: (from part (a)) (instantaneously at rest at point closest to wall) Figure 7.75 Thus The distance of the block from the wall is EVALUATE: The work done by F puts 5.0 J of mechanical energy into the system. No mechanical energy is taken away by friction, so the total energy at points B and C is 5.0 J.   7.79. IDENTIFY: SET UP: EXECUTE: . Use and for all the water that passes through the dam. is the volume of water in a height . (a) of water in the lake. (b) 90% of the stored energy is converted to electrical energy, so . . Change in level of the lake: . . EVALUATE: is much less than 150 m, so using the dam was a very good approximation.   . for all the water that passed through