Question

A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an

acceleration of magnitude a = 3.2m/s², as shown in (Figure 1).
Part A. What is the coefficient of kinetic friction between the block and the surface? Express your answer to two significant figures.
Part B. When the speed of the block slows to 0.40 m/s, is the magnitude of the acceleration greater than, less than, or equal to 3.2 m/s^2

Answer 1

μ = 0.33

Equal to 3.2 m/s²

Step-by-step explanation:

Draw a free body diagram of the block. There are three forces:

Normal force N pushing up.

Weight force mg pulling down.

Friction force Nμ pushing opposite the direction of motion.

Sum of forces in the y direction.

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction.

∑F = ma

Nμ = ma

Substitute.

mgμ = ma

μ = a/g

μ = (3.2 m/s²) / (9.8 m/s²)

μ = 0.33

As found earlier, the acceleration is a = gμ. Since g and μ are constant, a is also constant, so it does not change with velocity.