Question 1

A man pushes a box horizontally across a frictionless floor with a force of 50 N. The box moves a distance of 8 m in the direction of the applied force. How much work is done on the box?

Accepted Answer

Work is defined as the product of force and displacement in the direction of the force. Using the formula W = F × s × cos(θ), where θ is the angle between force and displacement. Since the force is applied horizontally and the box moves horizontally, θ = 0°, so cos(0°) = 1. Therefore, W = 50 N × 8 m × 1 = 400 J. The work done on the box is 400 joules. Option B is correct.

Question 2

A man pushes a box of mass 20 kg across a horizontal floor with a force of 100 N applied at an angle of 60° below the horizontal. The box moves 5 m. Calculate the work done by the applied force. (cos 60° = 0.5)

Accepted Answer

Work is defined as W = F·s·cosθ, where F is the applied force, s is the displacement, and θ is the angle between the force and displacement vectors. Here, the force is applied 60° below the horizontal, so the angle between the force direction and the horizontal displacement is 60°. Using the formula: W = 100 N × 5 m × cos(60°) = 100 × 5 × 0.5 = 250 J. This is an NCERT Class 9 concept where work depends on the component of force in the direction of motion. The vertical component of the force does no work on horizontal displacement because it is perpendicular to the motion. Therefore, answer is B.

Question 3

A 2 kg block is lifted vertically from the ground to a height of 5 m. The same block is then pushed horizontally along a frictionless surface for a distance of 10 m. If g = 10 m/s², what is the total work done against gravity during both operations?

Accepted Answer

The question tests understanding of work done by gravity and the directional nature of work. Work is defined as W = F·s·cosθ, where θ is the angle between force and displacement. During vertical lifting, the gravitational force (acting downward) opposes the upward displacement, so θ = 180°, giving W = mgh = 2 × 10 × 5 = 100 J. During horizontal motion, gravity acts perpendicular to displacement (θ = 90°), so cosθ = 0 and work done against gravity is zero. Total work done against gravity = 100 + 0 = 100 J. This is an NCERT Class 9 concept: work depends on the component of force in the direction of displacement. The horizontal motion contributes nothing because gravity is perpendicular to the motion. Therefore, the answer is B.

CAPF AC Work, Energy & Power

Work, Energy & Power— Rules & Concept

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Work, Energy & Power — Practice Questions

60-Second Revision — Work, Energy & Power