Given that a force →F acts on a body for time t, and displaces the body by →d. In which of the following cases, the speed of the body must not increase?
|→F|<|→d|
→F⊥→d
→F=→d
|→F|>|→d|
A
|→F|>|→d|
B
|→F|<|→d|
C
→F⊥→d
D
→F=→d
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Solution
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Work done by a force W=Fdcosθ
where F is the force acting on the body, d is the displacement of the body and θ is the angle between F and d.
If force is perpendicular to the displacement i.e. →F⊥→S, then we get θ=90o
We know cos90o=0
⟹W=0
Now from work energy theorem, change in kinetic energy of body ΔK.E=W
⟹ΔK.E=0 which implies that velocity remains the same.
Hence velocity will remain same when force is perpendicular to the direction of displacement
i.e. ^F⊥^d.
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