A crate of mass 5\text{ kg} is pulled from rest along a smooth inclined plane tilted at 30… — JAMB UTME

A crate of mass $5\text{ kg}$ is pulled from rest along a smooth inclined plane tilted at $30^\circ$ to the horizontal by a constant force of $40\text{ N}$ acting parallel to the slope. What is the kinetic energy of the crate in Joules after moving a distance of $6\text{ m}$ along the incline? (Take $g = 10\text{ m s}^{-2}$ )

Answer: 90 J

Answer

The final kinetic energy of the crate is

90\text{ J}

The kinetic energy gained equals the net work done on the crate. The total work put in by the pulling force is

40 \times 6 = 240\text{ J}

, while the gravitational potential energy gained is

5 \times 10 \times (6 \sin 30^\circ) = 150\text{ J}

. Subtracting the potential energy gained from the total work gives a net kinetic energy of

240 - 150 = 90\text{ J}

Step-by-Step Solution

Calculate the total work done by the applied force parallel to the incline

W_{\text{applied}} = 40\text{ N} \times 6\text{ m} = 240\text{ J}

Work done by a force is equal to force multiplied by displacement in the direction of the force.

Calculate the work done against gravity (potential energy gained)

\Delta PE = m g d \sin(30^\circ) = 5 \times 10 \times 6 \times 0.5 = 150\text{ J}

The height gained along an incline of length

d

and angle

\theta

h = d \sin\theta

Determine the net work done to find the final kinetic energy

KE = W_{\text{applied}} - \Delta PE = 240\text{ J} - 150\text{ J} = 90\text{ J}

According to the work-energy theorem, the net work done on an object equals its change in kinetic energy.

Key Concept

Work-Energy Theorem on an Inclined Plane

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