A block of mass 2.0 kg is pushed down an inclined plane of inclination 37o with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 m/s2. If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take g = 10 m/s2.
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