Question

A force of 44.0 N accelerates a 5.0-kg block at 6.3 m/s2 along a horizontal surface.

How large is the frictional force?
and What is the coefficient of friction?

Answer 1

Question#18927

A force of $44.0\,\mathrm{N}$ accelerates a 5.0-kg block at $6.3\,\mathrm{m/s^2}$ along a horizontal surface.

How large is the frictional force?

and What is the coefficient of friction?

Solution:

Let:

$$\begin{array}{l} F = 44\,\mathrm{N} \\ m = 5\,\mathrm{kg} \\ a = 6.3\,\mathrm{m/s^2} \\ \end{array}$$

Ffriction-? , k-?

$$\begin{array}{l} F - F_{\text{friction}} = ma \\ F_{\text{friction}} = F - ma \\ F_{\text{friction}} = 44 - 5 * 6.3 = 12.5\,\mathrm{N} \\ \end{array}$$

Such as: $F_{\text{friction}} = kmg, k = \frac{F_{\text{friction}}}{mg}$

$$k = \frac{12.5}{5 * 9.8} = 0.2551$$

Answer: the frictional force is $12.5\,\mathrm{N}$, the coefficient of friction is 0.2551.