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


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




=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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

A rectangular boxPhysics - Exemplar

A helicopter of mPhysics - Exemplar

There are four foPhysics - Exemplar

When a body <spanPhysics - Exemplar

Block <span lang=Physics - Exemplar

A block <span lanPhysics - Exemplar

Two <span lang="EPhysics - Exemplar