Answer to Question #94919 in Mechanics | Relativity for Rajan kumar

Question

an object of mass 20 kg is pulled out with the force of 1:50 newton on an inclined plane for the distance of 3 metre the plane mai an angle of 30 degree with the horizontal and the coefficient of kinetic friction between the object and the plane is 0.3 draw the free body diagram and calculate the work done by each force.

Answer 1

Solution:

The free body diagram:

The formula of the work done by the force is:

$W=FScos\alpha$

Let's find the work done by the force F:

$W_F=FScos(0^\circ)=150\cdot{3}=450$ J.

The work done by the friction force:

$W_{Fr}=F_fScos(180^\circ)=-\mu{}mgScos\theta=$

$=-0.3\cdot{}20\cdot{}9.8\cdot{}3\cdot{}cos(30^\circ)=-153$ J.

The work done by the force of gravity:

$W_{mg}=mgScos(90^\circ+\theta)=-mgSsin(\theta)=$

$=-20\cdot{}9.8\cdot{}3\cdot{}sin(30^\circ)=-294$ J.

the work done by the force N:

$W_N=NScos(90^\circ)=0$

Answer:

W_F = 450 J, W_Fr = -153 J, W_mg = -294 J, W_N = 0.