Question

A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and onto a floor where friction causes it to stop a distance D from the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is μk. 1)What is the macroscopic work done on the block by friction during this process? mgH

Answer 1

Answer:

$\displaystyle W=-m.g.H$

Explanation:

Energy Transformation

Also called as energy conversion, is the process where energy changes from one form to another. There are three types of energy present in this problem. When the object is at rest at the top of the ramp, it has gravitatory potential energy, calculated as

$U=m.g.H$

When the object slides down the frictionless ramp, it loses all of the potential energy since it's converted to kinetic energy, given by

$\displaystyle K=\frac{m.v^2}{2}$

Thus

$K=m.g.H$

Finally, when the object is in contact with a rough surface, all the energy is transformed into thermal energy. The work done by the friction force is equivalent to the change of kinetic energy, since all the velocity is lost during the process:

$\displaystyle W=\Delta E=K_f-K=0-K=-\frac{m.v^2}{2}$

Since the kinetic energy is equal to the original potential energy:

$\boxed{\displaystyle W=-m.g.H}$

The negative sign indicates the work was against the movement, i.e. the force and the displacement are 180° apart.

Note this result does not depend on the distance D needed to stop the block or the coefficient of kinetic friction.