Question

A box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration$\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$directed down the ramp. The ramp makes an angle of$\mathbf{40}\mathbf{°}$with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?

Short answer

${\mu }_k=0.47$

Step-by-step solution

Given data

• The angle of inclination of the ramp, $\theta =40°$.
• The acceleration down the ramp is $a=0.75\mathrm{m}/{\mathrm{s}}^2$.

Understanding the concept

The problem deals with Newton’s second law of motion, which states that the acceleration of an object can be written in terms of the net force acting upon the object and the mass of the object.Use Newton’s law of motion to calculate the kinetic friction.

Formula:

$$\begin{gathered} F=ma \\ {F}_N=mg\cos \theta \end{gathered}$$

Calculate the coefficient of kinetic friction between the box and the ramp

The normal force on the block on the inclined ramp is:

$F_N=mg\cos \theta$

To calculate acceleration, use Newton’s second law of motion:

$$\begin{gathered} mg\sin \theta -f_k=ma \\ mg\sin \theta -{\mu }_{k}mg\cos \theta =ma \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \left(9.8m/s^2\right)\times \sin \left(40°\right)-{\mu }_k\times \left(9.8m/s^2\right)\times \cos \left(40°\right)=0.75m/s^2 \\ {\mu }_k=0.74 \end{gathered}$$

Thus, the coefficient of kinetic friction is$0.74$.