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प्रश्न
A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is (i) in the direction of force, (ii) at an angle of 60° with the force, and (iii) normal to the force. (g = 10 N kg-1)
उत्तर
Force acting on the body = 10 kgf = 10 × 10 N = 100 N
Displacement, S = 0.5 m
Work done = force x displacement in the direction of force
(i) W = F × S
W = 100 × 0.5= 50 J
(ii) Work = force x displacement in the direction of force
W = F × S cosθ
W = 100 × 0.5 cos60o
W = 100 × 0.5 × 0.5(cos 60o = 0.5)
W = 25 J
(iii) Normal to the force:
Work = force x displacement in the direction of force
W = F × S cosθ
W = 100 × 0.5 cos90o
W = 100 × 0.5 × 0 = 0 J(cos90o = 0)
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