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A block of weight 100N is pushed by a force F on a horizontal 1 m/s2, when force is doubled its acceleration becomes 10 m/s2. The coefficient of friction is _(g = 10ms-2)(a) 0.2(b) 0.4(C) 0.6(d) 0.8 |
| Answer» GIVEN that,Weight, W=100NAcceleration, a=1m/s 2 Acceleration when FORCE is DOUBLE, a=10m/s 2 Now, Let the External Force be F and friction force be f.Then, F−f s =ma F−μN=ma F−μ100=10×1 F−μ100=10....(I)We know, When Force is doubled,Then,2F−f s =ma2F−μN=ma2F−μ100=10×102F−μ100=100....(II)Now, from equation (I) and (II)F=90NNow, put the value of F in equation (I)90−μ100=10−μ100=10−90μ= 10080 μ=0.8Hence, the COEFFICIENT of restitution is 0.8 | |