Chapter 4, Problem 64P

To determine

The coefficient of kinetic friction between the slide and the child.

Answer to Problem 64P

Solution:

Coefficient of kinetic friction=0.51

Explanation of Solution

Given:

Angle of inclination, $\theta ={34}^{\text{o}}$

Formula used:

From Newton’s second law of motion:

  ${\displaystyle \sum F=ma}$

Where, m is the mass and a is the acceleration.

Third equation of motion is:

  $V^2=V_o{}^2+2ad$

Where, V is the final velocity, $V_o$ is the initial velocity, a is the acceleration and d is the displacement.

Component of weight along the inclined plane is $mg\sin \theta$ .

Component of the weight perpendicular to the inclined plane is $mg\text{cos }\theta$ .

Frictional force is:

  $f=\mu N$

  $\mu$ is the coefficient of friction and N is the normal reaction force.

  $N=mg\text{cos θ}$

Calculation:

For frictionless inclined plane:

  $\begin{array}{l}mg\sin \theta =ma\\ a=g\sin \theta \\ V^2=V_o{}^2+2ad\\ V^2=0+2(g\sin \theta )(d)\\ V^2=2(g\sin \theta )(d)--------(1)\end{array}$

For the inclined plane with friction:

  $\begin{array}{l}mg\sin \theta -{\mu }_{k}mg\cos \theta =ma\\ a=g\sin \theta -{\mu }_{k}g\cos \theta \\ V_f{}^2=V_o{}^2+2ad\\ V_f=\frac{V}{2}\\ {\left[\frac{V}{2}\right]}^2=0+2(g\sin \theta -{\mu }_{k}g\cos \theta )(d)\\ {\frac{V}{4}}^2=2(g\sin \theta -{\mu }_{k}g\cos \theta )(d)------(2)\end{array}$

Substitute the value from equation (1) into equation (2) and solve.

  $\begin{array}{l}\frac{2(g\sin \theta )(d)}{4}=2(g\sin \theta -{\mu }_{k}g\cos \theta )(d)\\ \frac{\sin \theta }{4}=\sin \theta -{\mu }_k\cos \theta \\ {\mu }_k\cos \theta =\sin \theta -\frac{\sin \theta }{4}\\ {\mu }_k\cos \theta =\frac{3}{4}\sin \theta \\ {\mu }_k=\frac{3}{4}\tan \theta \\ {\mu }_k=\frac{3}{4}\tan {34}^{\text{o}}\\ {\mu }_k=0.51\end{array}$

Conclusion:

Thus, the coefficient of the kinetic friction between the slide and the incline is 0.51.