Question

Use the work energy theorem to rank the final kinetic energy of a ball based on the initial kinetic energy Ki, the magnitude of a constant force F on the ball, the displacement of the ball, d and the angle, theta between the displacement of the ball and the net force on the ball. Rank from greatest kinetic energy (1) to least kinetic energy (4).

1. ki= 150 J, F= 10 N, d=15 m, theta= 90 degrees
2. Ki= 300 J, F= 200 N, d= 1.5 m, theta= 180 degrees
3. ki= 200 J, F= 25 N, d= 4 m, theta= 0 degreee
4. ki= 450 J, F= 15 N, d= 30 m, theta= 150 degrees
4.​

Answer 1

150 J

-300 J

300 J

256.8 J

Step-by-step explanation:

Use the following equation: W = ΔK = Kf - Ki

Calculations below:

Ki = 150 J, F = 10 N, d = 15 m, θ = 90°

The work done on the ball is:

W = Fd cos θ = 10 × 15 × cos 90° = 0 J

Therefore, the final kinetic energy is:

Kf = Ki + W = 150 + 0 = 150 J

Ki = 300 J, F = 200 N, d = 1.5 m, θ = 180°

The work done on the ball is:

W = Fd cos θ = 200 × 1.5 × cos 180° = -600 J

Therefore, the final kinetic energy is:

Kf = Ki + W = 300 - 600 = -300 J

Ki = 200 J, F = 25 N, d = 4 m, θ = 0°

The work done on the ball is:

W = Fd cos θ = 25 × 4 × cos 0° = 100 J

Therefore, the final kinetic energy is:

Kf = Ki + W = 200 + 100 = 300 J

Ki = 450 J, F = 15 N, d = 30 m, θ = 150°

The work done on the ball is:

W = Fd cos θ = 15 × 30 × cos 150° = -193.2 J

Therefore, the final kinetic energy is:

Kf = Ki + W = 450 - 193.2 = 256.8 J