PHYS 2133 Chapter 6 Solutions

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Solutions

1

Work = force times displacement = 5.00 x 10³N x 3.00 x 10³m = 1.50 x 10⁷J.

2

a) The work done by the force is (150 N times 6 m) or 900 J.
b) Since the kinetic energy is not changing, according to the work-energy theroem the net force must be zero. Since the net force is zero, kinetic friction must be exactly 150 N.
Using kinetic friction = coefficient of friction times normal force we solve for m_k to find:
m_k = 150 N / (40 kg x 9.81 m/s/s) = 0.382

3

a) The kinetic energy is 0.5 mv² = 90.0 J.
By the work-energy theorem, the work done by the net force must be responsible for this gain of 90.0 J of kinetic energy.
Force times displacement = 90.0 J with the displacement of 0.500 m, the force must be 180 N.

4

Using the conservation of energy, the kinetic energy of the vaulter running down the ramp will equal the kinetic energy + potential energy at the top of the vault.
0.5 x 54 kg x (10 m/s)² = 0.5 x 54 kg x (1 m/s)² + 54 kg x 9.81 m/s/s x height. (Note that the only velocity at the top will be horizontal, the vertical component must be zero.)
Solving for the height, we find h = 5.05 m.
You should notice that the mass of the vaulter does not matter, it appears in all terms of the equation and will cancel in the solution.

5

a) The work done is equal to the change in kinetic energy, and is found to be 7.50 x 10⁴J.
b) Average power is work divided by time and is 2.5 x 10⁴ W.
c) Assuming the force accelerating the car is constant and can be found from Newton's second law to be 5000 N, we can find the instantanous power from the relations power = force x velocity. (note the assumption of constant force and instantanous velocity)
Therefore, power = 5000 N x 6.67 m/s = 3.33 x 10⁴W.

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