Q. 17

# A block of mass 2 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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