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1) A particle of mass m at rest is acted upon by a force F for a 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} } }



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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

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

7 4 7
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