Question

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 ?

Answer 1

`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.

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`