A force of 5N, making an angle θ with the horizontal, acting on an object and displaces it by 0.4m along the horizontal direction. If the work done by the force is 1J, then the horizontal component of the force is
A
1.5N
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B
2.5N
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C
3.5N
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D
4.5N
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Solution
The correct option is B2.5N
Here, the horizontal component of the applied force is Fcosθ
Work done by force on the body = 1J ⇒Fcosθ(S)=1J ∴Fcosθ=10.4N=2.5N