Section 4.1  Work

1. (Exercises 3 and 4) Calculate the work done and the power used when a force of 649 N pushes against the bottom of a 66.2-kg crate and lifts it to a height of 4.55 m above its starting position in 10.6 seconds.

In this case the force is vertical (upward) and the distance through which the force is exerted is also vertical, so 100% of the force is involved in doing the work.

W = F d = 649 N (4.55 m) = 2950 J

We could also find the work using change in potential energy. This problem will then pair with exercises 11 and 12 as well.

W = E_pF - E_pI = m g h_F - 0 = 66.2 kg (9.80 m/s²) (4.55 m) = 2950 J

Since h_I at ground level is zero, E_PI is therefore zero, as indicated above. Notice that both methods of calculating work resulted in the same numerical solution, so either method could have been used.

Section 4.2  Kinetic Energy and Potential Energy

2. (Exercises 11 and 12) How much work must be done to get a 0.025-kg golf ball to a height of 12.0 m above the tee (ground level) if the ball is traveling at 45 m/s when it reaches this height? Here we are finding work from the change in energy and, in this case, both the kinetic energy and the potential energy change, so we must use both.

W = E_P + E_k = [m g h_F - 0] + [¹/₂ m v_F ² - 0]

W = [0.025 kg (9.80 m/s²)(12.0 m)] + [¹/₂ (0.025 kg)(45 m/s)²]

W = 2.94 J + 25.3 J = 28.2 J

Section 4.3  Conservation of Energy

3. (Exercises 19 and 20) A child on a skateboard starts from rest at the top of a wooden track that is 8.0 m above ground level. What will be the child's speed when he reaches a point 1.5 m in vertical height from the bottom of the track? The mass of the child, along with his clothing and equipment, is 35 kg.

Using conservation of mechanical energy for the skater at the two levels on the track given above we have:

(E_k + E_p)_F = (E_k + E_p)_I

¹/₂ m v_F² + m g h_F = ¹/₂ m v_I² + m g h_I

Because the child started from rest at the top of the track, the initial speed is zero, so the kinetic energy at 1.5 m above the ground will be equal to the difference in potential energies.

¹/₂ m v_F² = m g h_I - m g h_F

Solving for v_F² in the previous equation gives: v_F² = 2 m g (h_I - h_F) / m = 2 g (h_I - h_F)

Notice that the m cancels, leaving only:

v_F² = 2 (9.8 m/s²) (8.0 m - 1.5 m) = 2 (9.80 m/s²) (6.5 m) = 127 m²/s² Taking the square root of this number gives: v_F = 11 m/s

Section 4.4  Power

4. (Exercises 25 and 26) A 48-kg girl climbs a vertical ladder to a height of 2.5 m above the ground in 15 seconds.

(a) How much work has she done?
(b) What power output must she exert during the climb?

Solutions:

(a) W = F d = m g h = 48 kg (9.80 m/s²) ( 2.5 m) = 1200 J

(b) P = W / t = 1200 J / 15 s = 80 W

If you prefer, you can express the power in "horsepower."

P = 80 W (1 hp / 746 W) = 0.11 hp

Return to Previous Page