A constant horizontal force acts on a body of mass 10\text{ kg}, accelerating it from rest… — JAMB UTME

A constant horizontal force acts on a body of mass $10\text{ kg}$ , accelerating it from rest to a speed of $12\text{ m s}^{-1}$ in a time of $4\text{ s}$ along a smooth horizontal surface. What is the average power delivered by the force during this time interval?

Answer: 180 W

Answer

The average power delivered by the force during the 4-second interval is

180\text{ W}

By the work-energy theorem, the total work done by the constant force equals the gain in kinetic energy:

W = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times 144 = 720\text{ J}

. The average power is the rate at which work is performed over time:

P = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

Step-by-Step Solution

Calculate the final kinetic energy acquired by the body.

E_k = \frac{1}{2} m v^2 = \frac{1}{2} (10\text{ kg})(12\text{ m s}^{-1})^2 = 720\text{ J}

Since the body starts from rest on a smooth surface, all work done by the net force goes into increasing its kinetic energy.

Divide the total work done by the elapsed time to find the average power.

P_{\text{avg}} = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

Average power is defined as the rate of doing work over a given time interval.

Key Concept

Work-Energy Theorem and Average Power

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