Answer :
m=mass of the block = 2kg
θ=angle of inclination = 37°
Force applied on the box = F = 20N
Acceleration produced in the box = 10m/s2
(a.) We have to calculate work done by the applied force in the first second, i.e. t=1sec
We know, here the work done W=Fs
(Actually W=FscosΦ but cosΦ is eventually equal to1 as Φ, the angle between force F and displacement s, is zero)
For W, we need to find s.
Using equations of motion, s= ut+at2/2
where, s=distance travelled
t=time taken
u=initial speed
a=acceleration
⇒ s= 0×1+10×1×1/2
⇒ s= 5m
∴ Work done by applied force W = 20×5 = 100 J
(b.) Work done by the weight of the box will be equal to the potential energy of the box.
W = Potential energy of the box
= mass×g×height
=mgh
=m×g×s×sin37°
=2×10×5×sin37°
=60 J
(c.) From the free body diagram, we can see frictional force = μR = mgsinθ
Work done by this force on the body = W = mgsinθ × s × cosΦ
(Φ=Angle between displacement s and frictional force)
W = 2×10×sin37°×s×cos180°
W = -60 J
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