Question

A 10 kg box is being pushed by a 20N force. what is the coefficient of kinetic friction?

Answer 1

Answer on Question 72584, Physics, Mechanics | Relativity

Question:

A 10 kg box is being pushed by a 20 N force. What is the coefficient of kinetic friction?

Solution:

By the definition of the coefficient of kinetic friction we have:

$$\mu_{k} = \frac{F_{appl}}{W} = \frac{F_{k}}{N},$$

here, $F_{appl}$ is the horizontal pushing force, $W = mg$ is the weight of the box directed downward, $F_{k}$ is the kinetic friction force directed opposite to the horizontal pushing force and equal to it, $N$ is the force of reaction directed upward and equal to the weight of the box.

From this formula we can find the coefficient of kinetic friction:

$$\mu_{k} = \frac{F_{appl}}{W} = \frac{F_{appl}}{mg} = \frac{20\,N}{10\,kg \cdot 9.8\, \frac{m}{s^{2}}} = 0.2.$$

Answer:

$$\mu_{k} = 0.2.$$

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