Question

1) A particle of mass m at rest is acted upon by a force F for time t. Its kinetic energy after an interval t is: {answer should come $\frac{ F^{2} t^{2} } {2m}$}2) A rod of length 1 m and mass 0.5 kg hinged at one end, is initially hanging vertical. The other end is now raised slowly until it makes an angle $60^{o}$ with the vertical. The required work is : {use g = 10 m/$s^{2}$}(answer should come $\frac{5}{4} J$)3) A body is dropped from a certain height. When it loses U amount of it energy it acquires a velocity 'v'. The mass of the body is :     {answer should come $\frac{2U}{ v^{2} }$}

Answer 1

Answer:

Uniform force F acts during t seconds,

       the change in momentum is = impulse = F * t.

    Initial momentum = 0  => final momentum p = Ft

Kinetic energy = p²/2m = F² t² / 2 m

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initial position of center of mass from the hinge = 1/2 meter

the change in height of center of mass = 1/2 - 1/2 Cos 60 = 1/4 meter

change in potential energy = m g h = 0.5 kg * 10 m/s² * 1/4 m = 5/4 Joules

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gain in kinetic energy =  1/2 m v² = U  = lost energy

     m = 2 U / v²