A force is inclined at 60∘ to the horizontal produces a displacement of →s=(^i+^j)m. If the component of force in horizontal direction is 50N, the net work done by the force on the body is
A
−37J
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B
37J
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C
−137J
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D
137J
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Solution
The correct option is D137J Given, Horizontal component of force (Fx)=50N
Displacement of the body (→s)=(^i+^j)m
Since, Fx=Fcosθ=50N and θ=60∘ F=50cos60∘=100N
Vertical component of force (Fy)=Fsinθ=100sin60∘=50√3N
So, →F=(Fx^i+Fy^j)=(50^i+50√3^j)
Net work done (W)=→F.→s W=(50^i+50√3^j).(^i+^j) W=50+50√3, as √3≈1.73 W=137J