A rope pulls on the lower block in the figure with a tension force of 20 N.The coefficient of kinetic friction between the lower block and the surface is 0.16.The coefficient of kinetic friction between the lower block and the upper block is also 0.16.The pulley has no appreciable mass or friction.What is the acceleration of the 2.0 kg block? A) 4.1 m/s² B) 5.1 m/s² C) 8.4 m/s² D) 9.2 m/s²

Question

Quizplus · Accepted Answer

First, we need to find the frictional force acting on the lower block. 
Frictional force = coefficient of kinetic friction * normal force 
Using the free-body diagram of the lower block, we can see that:
Normal force = Weight of lower block - Tension force = mg - 20, where g is the acceleration due to gravity (9.8 m/s²) and m is the mass of the lower block (2.0 kg).
Therefore, Frictional force = 0.16 * (mg - 20)
Next, we need to use the free-body diagram of the upper block to find the net force acting on it. We can see that:
Net force = Tension force - Frictional force
Using Newton's second law (F = ma), we can write:
(Weight of upper block - Tension force) - (0.16 * Tension force) = (mass of upper block) * (acceleration of blocks)
Simplifying and substituting given values, we get:
(5g - 20) - (0.16 * 20) = 2a
5g - 32 = 2a
a = (5g - 32) / 2
a ≈ 4.1 m/s²

A Rope Pulls on the Lower Block in the Figure