Question

A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient of kinetic friction is 0.25 . as the block goes 2.61 m down the plane the mechanical energy of the earth block system chenges by ?

Answer 1

Answer:

Change in mechanical energy = work done by friction

so it is equal to

$W = -8.16 J$

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

$W = F.d$

here we have

$F = \mu F_n$

here we know that

$F_n = mg cos\theta$

so we have

$F_n = 2.55(9.81)(cos60)$

$F_n = 12.5 N$

Now the friction force on the block is given as

$F_f = \mu F_n$

$F_f = 0.25 \times 12.5$

$F_f = 3.13 N$

now work done by the friction is given as

$W = -(3.13)(2.61)$

$W = -8.16 J$