Question

A horizontal force of magnitude 46.7 N pushes a block of mass 4.35 kg across a floor where the coefficient of kinetic friction is 0.635. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 2.96 m across the floor? (b) During that displacement, the thermal energy of the block increases by 35.0 J. What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

Answer 1

(a) 138.2 J

Since the applied force is parallel to the displacement of the block, the work done by the force is given by:

$W=Fd$

where

F = 46.7 N is the magnitude of the force

d = 2.96 m is the displacement of the block

Substituting the numbers into the equation, we find

$W=(46.7 N)(2.96 m)=138.2 J$

(b) 45.1 J

In order to calculate the total energy dissipated among the floor and the block as thermal energy, we have to calculate the work done by the frictional force, which is

$W_f = F_f d = (-\mu mg)d$

where$\mu=0.635$ is the coefficient of friction

m = 4.35 kg is the mass of the block

g = 9.8 m/s^2

d = 2.96 m is the displacement

and the negative sign is due to the fact that the frictional force has opposite direction to the displacement.

Substituting, we find

$W_f =-(0.635)(4.35 kg)(9.8 m/s^2)(2.96 m)=-80.1 J$

The magnitude of this work is equal to the sum of the thermal energy dissipated between the floor and the block:

$W_f = E_{floor}+E_{block}$

And since we know

$W_f = 80.1 J\\E_{floor}=35.0 J$

we find

$E_{floor}=W_f-E_{block}=80.1 J-35.0 J=45.1 J$

(c) 58.1 J

According to the work-energy theorem, the increase in kinetic energy of the block must be equal to the work done by the applied force minus the work done by friction (which becomes wasted thermal energy):

$\Delta K=W-W_f$

Substituting

$W=138.2 J\\W_f = 80.1 J$

We find

$\Delta K=138.2 J-80.1 J=58.1 J$