A slide-loving pig slides down a certain${\mathbf{35}}^{\mathbf{\circ }}$slide in twice the time it would take to slide down a frictionless${\mathbf{35}}^{\mathbf{\circ }}$slide.What is the coefficient of kinetic friction between the pig and the slide?

Angle of the slide,$\theta ={35}^{\text{o}}$

Consider the time required to slide on frictionless surface is t.

So, the required time for the surface with friction is $t\text{'}=2t$

Use Newton's 2nd law of motion and kinematic equations of motion.According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\sum Ma$….. (i)

Here, F is the net force, mis mass and ais an acceleration.

$s=v_{0}t+\frac{1}{2}a{t}^2$…… (ii)

Case(1):(Consider surface is frictionless)

According to the Newton’s 2nd law of motion along x direction(pig is moving along x so consider its acceleration is a), the equation (i) can be written as,

$$\begin{gathered} \sum _{}^{}{F}_x=m{a}_x \\ {f}_s-Mg\sin \left(\theta \right)=Ma \end{gathered}$$

Since, surface is frictionless$f_s=0\text{N}$

Thus,$a=-g\sin \left(\theta \right)$

This is the acceleration of a pig on a frictionless surface.

Case(2):Consider the surface is frictional.

$$\begin{gathered} \sum _{}^{}{F}_x=m{a}_x \\ {f}_s-Mg\sin \left(\theta \right)=Ma\text{'} \\ {\mu }_k\left(Mg\cos \left(\theta \right)\right)-Mg\sin \left(\theta \right)=Ma\text{'} \\ a\text{'}={\mu }_{k}g\cos \left(\theta \right)-g\sin \left(\theta \right) \end{gathered}$$

This is the acceleration of a pig on a friction surface.

Distance is constant in both cases i.e. with friction and without friction is same so by usingequation (ii),

$s=v_{0}t+\frac{1}{2}a{t}^2$

Distance travelled on frictionless surface is same as Distance travelled on frictional surface

$v_{0}t+\frac{1}{2}a{t}^2=v_{0}t\text{'}+\frac{1}{2}a\text{'}t{\text{'}}^2$

Since the pig is at rest initially, the above equation can be written as,

$$\begin{gathered} \left(0\right)t+\frac{1}{2}a{t}^2=\left(0\right)\left(2t\right)+\frac{1}{2}a\text{'}(2t{)}^2 \\ \frac{1}{2}a{t}^2=\frac{1}{2}a\text{'}(2t{)}^2 \\ a=4a\text{'} \\ -g\sin \left(\theta \right)=4\left({\mu }_{k}g\cos \left(\theta \right)-g\sin \left(\theta \right)\right) \end{gathered}$$

Substitute the values and solve as:

$\begin{array}{rcl}{\mu }_k& =& \frac{3\sin \left(\theta \right)}{4\cos \left(\theta \right)}\\ & =& 0.525\\ & & \end{array}$

Therefore, the coefficient of kinetic friction is$0.5251$