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A 40N force is applied on a 2 kg block placed onarough inclined plane having angle of inclination 60°with horizontal. If coefficient of friction is 0.4, find thework done by frictional force when the displacementof block is 2 m.to-F=40N(1) -6 Joule(3) +6 Jule(2) -12 Joule(4) -8 Joule |
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Answer» Normal reaction is = mgcos(60) = 2*10*1/2 = 10N now friction force will act opposite to F so, Friction force = μN = 10*0.4 = 4N so, work done for distance 2m , in opposite direction will be = -4N*(2) = -8J option 4 |
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