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 ?

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#