Question

A 95kg clock is pushed along the floor at constant speed by a force of 560N . Calculate the coefficient of friction.

Answer 1

Applying the Newton’s Second Law of Motion we get (since the clock is pushed along the floor at constant speed the acceleration is zero):

$$\sum {F_{x}} = ma_x = 0,$$ $$F_{push} - F_{fr} = 0,$$ $$F_{push} = F_{fr} = \mu N = \mu mg.$$

From this formula we can find the coefficient of friction:

$$\mu = \dfrac{F_{push}}{mg} = \dfrac{560N}{95kg \cdot 9.8 \dfrac{m}{s^2}} = 0.60$$

Answer:

$\mu = 0.60$