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A stone of 1 kg is thrown with a velocity of 20 m s^-1 acrossthe frozen surface of a lake and comes to rest after travellinga distance of 50 m. What is the force of friction between thestone and the ice? |
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Answer» m = 1kgu = 20m/s. v = 0m/s. s(distance travelled) = 50m Using third equation of motionv²=u²+2as0² = (20)²+2(a)(50)-400 = 100aa = -400/100 = -4m/s² (retardation) F = m×aF = 1×(-4) = -4N. (negative sign indicates the opposing force which is Friction) |
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