IB Physics SL GOHS
5.5-5.6 Worksheet: Conservation of Energy and Power
Conceptual Questions
16. The feet of a standing person of mass m exert a force equal to mg on the floor, and the floor exerts an equal
and opposite force upwards on the feet, which we call the normal force. During the extension phase of a vertical
jump (see page 145), the feet exert a force on the floor that is greater than mg, so the normal force is greater
than mg. As you learned in Chapter 4, we can use this result and Newton’s second law to calculate the acceleration
of the jumper: a = Fnet/m = (n - mg)/m. Using energy ideas, we know that work is performed on the jumper to give him or her kinetic energy. But the normal force can’t perform any work here, because the feet don’t undergo any displacement. How is energy transferred to the jumper?
20. Discuss the energy transformations that occur as a pole vaulter runs at high speeds and attempts to clear a bar
that is about 5 m from the ground. In your analysis, you must consider changes in the kinetic energy of the
runner, the elastic potential energy of the pole as it bends, and the gravitational potential energy of the vaulter. Ignore rotational motion.
MORE ON BACK!!!!!
Problems
48. A skier of mass 70 kg is pulled up a slope by a motordriven cable. (a) How much work is required to pull him 60 m up a 30° slope (assumed frictionless) at a constant speed of 2.0 m/s? (b) What power must a motor have to perform this task?
59. An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow? (b) How much work does the archer do in pulling the bow?
60. A block of mass 12.0 kg slides from rest down a frictionless 35.0° incline and is stopped by a strong spring with k= 3.00 x 104 N/m. The block slides 3.00 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?
ANSWERS
Practice F
Conceptual Questions
16. Work is actually performed by the thigh bone (the femur) on the hips as the torso moves upwards a distance h. The force on the torso is approximately the same as the normal force (since the legs are relatively light and are not moving much), and the work done by minus the work done by gravity is equal to the change in kinetic energy of the torso.
At full extension the torso would continue upwards, leaving the legs behind on the ground (!), except that the torso now does work on the legs, increasing their speed (and decreasing the torso speed) so that both move upwards together.
Note: An alternative way to think about problems that involve internal motions of an object is to note that the net work done on an object is equal to the net force times the displacement of the center of mass. Using this idea, the effect of throwing the arms upwards during the extension phase is accounted for by noting that the position of the center of mass is higher on the body with the arms extended, so that total displacement of the center of mass is greater.
20. Before the jump, the system (vaulter plus pole) has kinetic energy. After the vaulter leaves the ground and the pole is bent, the system has less kinetic energy, but the gravitational potential energy of the system increases and some energy is stored as elastic potential energy in the bent pole. When the vaulter reaches the top of the vault, the kinetic energy is at its minimum, the gravitational potential energy is at its maximum, and no energy is stored in the pole, which is now straight. Some mechanical energy is lost due to air resistance and the frictional force between the pole and the ground during the ascent.
Problems
5.48 (a) , but because the speed is constant. The skier rises a vertical distance of . Thus,
(b) The time to travel 60 m at a constant speed of is 30 s. Thus, the required power input is
5.59 (a) The equivalent spring constant of the bow is given by as
(b) From the work-energy theorem applied to this situation,
The work done pulling the bow is then
5.60 Choose at the level where the block comes to rest against the spring. Then, in the absence of work done by non-conservative forces, the conservation of mechanical energy gives
or . Thus,