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} } }

1

Answers

2015-01-13T10:35:03+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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