A 3.50 kg block is pulled along a moving conveyor belt at a constant speed of 0.500 m/s relative to a stationary observer while the belt moves at a constant speed of 0.200 m/s in the same direction. If the coefficient of kinetic friction is 0.400, the magnitude of the mechanical energy dissipated, in J, caused by the force of friction on the block in 8.00 s is
A) 5.6.
B) 22.0.
C) 32.9.
D) 54.8.
E) 76.8.

Question

Quizplus · Accepted Answer

First, we need to find the force of friction acting on the block. The formula for frictional force is:

$f_k = \mu_k N$

where $f_k$ is the kinetic frictional force, $\mu_k$ is the coefficient of kinetic friction, and $N$ is the normal force.

The normal force $N$ is equal to the weight of the block since it is on a horizontal surface. Thus,

$N = mg = (3.50 kg)(9.81 m/s^2) = 34.335 N$

And so,

$f_k = (0.400)(34.335 N) = 13.734 N$

The work done by friction over a distance $d$ is given by:

$W_f = f_k d$

Here, we can use the fact that the block moves at a constant speed relative to the observer. We can assume that the block stays in the same position relative to the conveyor belt (i.e. it is not slipping) and therefore moves with the belt. Thus, over a time interval of $t = 8.00 s$, the block travels a distance of:

$d = v_{b,rel} t = (0.500 m/s - 0.200 m/s)(8.00 s) = 2.40 m$

And so, the work done by friction is:

$W_f = (13.734 N)(2.40 m) = 32.9616 J$

Therefore, the magnitude of the mechanical energy dissipated by friction in 8.00 s is 32.9 J, and the best choice is (C).

A 350 Kg Block Is Pulled Along a Moving Conveyor Belt