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Question

Physics Question on work, energy and power

A body moves a distance of 10m10\, m along a straight line under a action of 5N5\, N force. If work done is 25J25 \,J, then angle between the force and direction of motion of the body will be

A

7575^{\circ}

B

6060^{\circ}

C

4545^{\circ}

D

3030^{\circ}

Answer

6060^{\circ}

Explanation

Solution

Component of force in the direction of displacement should be taken.
Work is measured by the product of the applied force and the displacement of the body in the direction of the force

Work == Force ×\times displacement
W=(Fcosθ)×ΔsW=(F \cos \theta) \times \Delta s
Given, W=25J,F=5N,Δs=10mW=25\, J, F=5\, N, \Delta s=10 \,m
cosθ=WFΔs\therefore \cos \theta=\frac{W}{F \cdot \Delta s}
=255×10=12=\frac{25}{5 \times 10}=\frac{1}{2}
θ=cos1(12)=60\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}
Hence, angle between force and direction of body is 6060^{\circ}.