Question

A 6 kg box has a coefficient of friction with a table of 0.7, and it’s slowing down across the table with no pushing force. What is the acceleration of the box?

A. 42 m/s^2
B. 7 m/s^2
C. 6 m/s^2
D. 0.7 m/s^2

Answer 1

B. 7 m/s²

Step-by-step explanation:

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

Normal force N pushing up.

Weight force mg pulling down.

Friction force Nμ pushing against 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

-mgμ = ma

a = -gμ

a = -(9.8 m/s²) (0.7)

a ≈ -7 m/s²

The magnitude of the box's acceleration is 7 m/s².