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 opposite 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

The mechanical energy dissipated due to friction is calculated using the work done by the friction force over a distance. The friction force can be found using the formula $f = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. Since the block is moving at a constant speed, the normal force equals the weight of the block, $N = mg$, where $m = 3.50 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$.First, calculate the friction force:$$f = \mu_k mg = 0.400 \times 3.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 13.72 \, \text{N}$$The distance the block moves relative to the conveyor belt in 8.00 s is the relative speed multiplied by time. However, since the question asks for the energy dissipated due to friction, we need to consider the distance the block moves relative to a stationary observer. The speed of the block relative to the stationary observer is the sum of the speeds of the block and the conveyor belt (since they move in opposite directions), which is $0.500 \, \text{m/s} + 0.200 \, \text{m/s} = 0.700 \, \text{m/s}$.The distance moved in 8.00 s is:$$d = vt = 0.700 \, \text{m/s} \times 8.00 \, \text{s} = 5.6 \, \text{m}$$The work done by the friction force, which equals the mechanical energy dissipated, is:$$W = fd = 13.72 \, \text{N} \times 5.6 \, \text{m} = 76.832 \, \text{J}$$Rounding to the nearest tenth gives 76.8 J, which corresponds to choice E.

A 350 Kg Block Is Pulled Along a Moving Conveyor Belt