A force given by the relation F = 8 (N/s)t acts on
a body of mass 2 kg initially at rest on a smooth
surface where t is in seconds. Find the work done
by this force on the body during first 2 seconds
of its motion. Assume direction of force remain
constant and body moves in the direction of
force.
(a) 64 J (b) zero (c) - 64 J (d) 32 J
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Answer:Work done = Force*Distance*cos(theta)
As theta = 0° here, so cos(0°)= 1
Distance travelled in 2sec:
Acceleration = 8/2 = 4m/s^2
S= ut + 0.5*a*t^2 = 0 + 0.5*4*2*2 = 8m
Work done by force = FS = 8*8 =64J
Explanation:
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