GiancoliPhysics for Scientists & Engineers, 4th Edition
CHAPTER 8: Conservation of Energy
Solutions to Assigned Problems
8.The force is found from the relations on page 189.
19.Use conservation of energy. The level of the ball on the uncompressed spring is taken as the zero location for both gravitational potential energy and elastic potential energy It is diagram 2 in the figure. Take “up” to be positive for both x and y.
(a)Subscript 1 represents the ball at the launch point, and subscript 2
represents the ball at the location where it just leaves the spring, at the uncompressed length. We have and Solve for
(b)Subscript 3 represents the ball at its highest point. We have and Solve for
20.Since there are no dissipative forces present, the mechanical energy of the roller coaster will be conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero location for gravitational potential energy. We have and
Point 2:
Point 3:
Point 4:
27.The maximum acceleration of 5.0 g occurs where the force is at a maximum. The
maximum force occurs at the bottom of the motion, where the spring is at its maximum compression. Write Newton’s second law for the elevator at the bottom of the motion, with up as the positive direction.
Now consider the diagram for the elevator at various points in its motion. If there are no non-conservative forces, then mechanical energy is conserved. Subscript 1 represents the elevator at the start of its fall, and subscript 2 represents the elevator at the bottom of its fall. The bottom of the fall is the zero location for gravitational potential energy There is also a point at the top of the spring that is the zero location for elastic potential energy (x = 0). We have and Apply conservation of energy.
48.Note that the difference in the two distances from the center of the Earth, is the same as the height change in the two positions, Also, if the two distances are both near the surface of the Earth, then
49.The escape velocity for an object located a distance r from a mass M is given by Eq. 8-19, The orbit speed for an object located a distance r from a mass M is
(a)
(b)
Since the orbiting object will not escape the orbit.
52.(a)With the condition that at the potential energy is given by The
kinetic energy is found from the fact that for a circular orbit, the gravitational force is a centripetal force.
(b)As the value of E decreases, since E is negative, the radius r must get smaller. But as the radius gets smaller, the kinetic energy increases, since If the total energy decreases by 1 Joule, the potential energy decreases by 2 Joules and the kinetic energy increases by 1 Joule.
70.The force to lift the water is equal to its weight, and so the work to lift the water is equal to the
weight times the vertical displacement. The power is the work done per unit time.
85.(a)The tension in the cord is perpendicular to the path at all times, and so the tension in the cord
does not do any work on the ball. Thus only gravity does work on the ball, and so the mechanical energy of the ball is conserved. Subscript 1 represents the ball when it is horizontal, and subscript 2 represents the ball at the lowest point on its path. The lowest point on the path is the zero location for potential energy We have , and Solve for
(b)Use conservation of energy, to relate points 2 and 3. Point 2 is as described above. Subscript 3
represents the ball at the top of its circular path around the peg. The lowest point on the path is the zero location for potential energy We have and . Solve for
86.The ball is moving in a circle of radius If the ball is to complete the circle with the string just going slack at the top of the circle, the force of gravity must supply the centripetal force at the top of the circle. This tells the critical (slowest) speed for the ball to have at the top of the circle.
To find another expression for the speed, we use energy conservation. Subscript 1 refers to the ball at the launch point, and subscript 2 refers to the ball at the top of the circular path about the peg. The zero for gravitational potential energy is taken to be the lowest point of the ball’s path. Let the speed at point 2 be the critical speed found above.
If h is any smaller than this, then the ball would be moving slower than the critical speed when it reaches the top of the circular path, and would not stay in centripetal motion.
88.The spring constant for the scale can be found from the 0.5 mm compression due to the 760 N force. Use conservation of energy for the jump. Subscript 1 represents the initial location, and subscript 2 represents the location at maximum compression of the scale spring. Assume that the location of the uncompressed scale spring is the 0 location for gravitational potential energy. We have and Solve for which must be negative.
90.(a)Draw a free-body diagram for the block at the top of the curve. Since the
block is moving in a circle, the net force is centripetal. Write Newton’s second law for the block, with down as positive. If the block is to be on the verge of falling off the track, then
Now use conservation of energy for the block. Since the track is frictionless, there are no non-conservative forces, and mechanical energy will be conserved. Subscript 1 represents the block at the release point, and subscript 2 represents the block at the top of the loop. The ground is the zero location for potential energy We have and Solve for h.
(b)See the free-body diagram for the block at the bottom of the loop. The net
force is again centripetal, and must be upwards.
The speed at the bottom of the loop can be found from energy conservation, similar to what was done in part (a) above, by equating the energy at the release point (subscript 1) and the bottom of the loop (subscript 2). We now have and Solve for
(c)Again we use the free body diagram for the top of the loop, but now the normal force does not
vanish. We again use energy conservation, with and Solve for
(d)On the flat section, there is no centripetal force, and
91.(a)Use conservation of energy for the swinging motion. Subscript 1
represents the student initially grabbing the rope, and subscript 2 represents the student at the top of the swing. The location where the student initially grabs the rope is the zero location for potential energy We have and Solve for
Calculate the angle from the relationship in the diagram.
(b)At the release point, the speed is 0, and so there is no radial acceleration,
since Thus the centripetal force must be 0. Use the free-body diagram to write Newton’s second law for the radial direction.
(c)Write Newton’s second law for the radial direction for any angle, and solve for the tension.
As the angle decreases, the tension increases, and as the speed increases, the tension increases. Both effects are greatest at the bottom of the swing, and so that is where the tension will be at its maximum.
98.It is shown in problem 52 that the total mechanical energy for a satellite orbiting in a circular orbit of radius r is That energy must be equal to the energy of the satellite at the surface of the Earth plus the energy required by fuel.
(a)If launched from the equator, the satellite has both kinetic and potential energy initially. The
kinetic energy is from the speed of the equator of the Earth relative to the center of the Earth. In problem 53 that speed is calculated to be 464 m/s.
(b)If launched from the North Pole, the satellite has only potential energy initially. There is no
initial velocity from the rotation of the Earth.
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