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A force given by the relation F = 8t, (F in Newton and t in sec.) acts on a body of mass 2kg. Initially at rest. Choose the correct option.A. Power of force is continuous increasing.B. The work done by this force on the body during first 2 seconds of its motion is 64 J.C. Velocity at 2 seconds after beginning of motion is 8 m/s.D. Power of force is constant. |
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Answer» Correct Answer - A::B::C Let the both blocks are moving together, then `a = (60)/(40) = 1.5 m//s^(2)` thus, force on lower block has to be F = Ma = `30 xx 1.5 = 45N` Next `f_(L) = mu N = 0.5 xx 10 xx 10 = 50 N` Since `F lt f_(L)` , both the blocks will move together, thus force of friction = 45N |
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