Answer :

asa2.png


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


as3.png


(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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