Question

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 work, in J, done on the block by the force of friction

in 8.00 s is

Answer 1

Let's first find the velocity of the block with respect to belt:

$$v_{block}=0.5\ \dfrac{m}{s}-0.2\ \dfrac{m}{s}=0.3\ \dfrac{m}{s}.$$

Then, we can find the distance traveled by the block in 8.0 seconds:

$$d=v_{block}t=0.3\ \dfrac{m}{s}\cdot8.0\ s=2.4\ m.$$

Finally, we can find the work done on the block by the force of friction in 8.00 s:

$$W_{fr}=F_{fr}d=\mu_kmgd,$$ $$W_{fr}=0.4\cdot3.5\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot2.4\ m=32.9\ J.$$