A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction 0.1. compute the
a) work done by the applied force in 10seconds
b) work done by friction in 10 seconds
c)work done by net force on the body in 10 seconds
d) change in kinetic energy of the body in 10 seconds
and interpret your results.

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Answers

  • Brainly User
2014-11-21T21:48:01+05:30
Here, F=7 f=2(10)(0.1)=2 a=(7-2)/2=2.5. Displacement in 10 seconds, S=125. 1) F.S=7(125)=875. 2) f.S=-2(125)=-250. 3) (F-f).S=5(125)=625. 4) ΔK.E = Net work done = 625
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2014-11-23T10:27:43+05:30

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Initial speed = 0
Friction force = 0.1 * 2 kg * 10 m/sec² = 2 N,  opposite to applied force
Net force on the body = 7 N - 2 N = 5 Newtons
acceleration a = 5 N/ 2kg = 2.5 m/sec²
Distance traveled in 10 sec = 1/2 a t² = 125 m

1)  work done by applied force =  F . S = 125 * 7 = 875 Joules

2)  work done by  friction = F . S = - 2 * 125  = - 250 Joules

3) Work done by net force in 10 sec = F . S = 5 * 125 = 625 Joules
      This is also equal to work done by applied force + work done by friction

4) change in kinetic energy = work done by the net force = 625 J

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