A 2.2 kg block starts from rest on a rough inclined plane that makes an angle 25^@ with the horizontal. The coefficient of kinetic friction is 0.25. As the block goes 2 m down the plane, the mechanical energy of the Earth-block system changes by ?

1 Answer
Feb 8, 2018

9.96 J

Explanation:

Here,value of maximum frictional force acting is mumgcos 25(Note,here this amount of frictional force will act,as it's value is smaller than the downward component of the weight of the block i.e mg sin 25

So,work done by frictional force is mu mg cos 25 *2=4.98 *2=9.96 J

So,the this amount of energy will be lost from the total energy of the system.

Alternatively,

Suppose,energy lost due to frictional force is E

So,we can write,total energy 2m above the final point = kinetic energy at the final point +energy lost due to frictional force.

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Now.total energy 2m above the final point or at the point of journey was purely potential energy i.e mgx Now,from the diagram, x/l = sin 25 or. x = l sin 25=0.845m, so mgx =18.6J

Now,if at the final point it gains a velocity of v then we can use, v^2 = u^2 +2as

Here, u=0,s = 2 and a=acceleration=Total downward force acting/mass= (mg(sin 25-mu cos 25))/m = g(sin 25- mucos 25)

Putting the values we get, v^2=7.85

So,kinetic energy at final point =1/2m v^2 = 8.64 J

So,energy lost due to frictional force is (18.6-8.64)=9.96 J