A block of mass 2kg initially at rest moves under the action of an applied horizontal force of 6N on a rough horizontal surface. The coefficient of friction between block and surface is 0.1. The work done by the applied force in 10s is (Take g=10ms−2).
200J
600J
−200J
−600J
A
−600J
B
200J
C
−200J
D
600J
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Solution
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The various forces acting on the block is as shown in the figure. Here, m=2kg,μ=0.1,F=6N,g=10ms−2
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Q5
A body of
mass 2 kg initially at rest moves under the action of an applied
horizontal force of 7 N on a table with coefficient of kinetic
friction = 0.1.
Compute
the
(a) work
done by the applied force in 10 s,
(b)
work done by friction in 10 s,
(c)
work done by the net force on the body in 10 s,