Answer to Question #146410 in Classical Mechanics for Shreya Sehgal

Question

A car starts from rest at the top of a hill.
The car rolls frictionlessly down the hill. The hill is sloped at an angle
of theta and the length of the hill is x1. At the bottom of the hill, the car
transitions to level ground and immediately slams on the brakes. The
car skids to a stop over some unknown distance (labeled as x2). The
coefficient of kinetic friction between the car and the ground on the
level section is mu k. Suppose that the length of the hill is equal to half
the distance that the car slides before coming to a stop. What value
for the coefficient of kinetic friction would allow this to happen?
Answer in terms of theta.

Answer 1

Acceleration of the car along the inclined surface $(a)=g\sin\theta$

Let initial speed of the car $u=0$

final speed of the car $=v$

Now, $v^2=u^2+2ax_1$

$v^2=0+2g\sin\theta x_1$

$v=\sqrt{2g\sin\theta x_1}$

Let retardation of the car at the ground $(a_1)=\mu_k g$

and $x_2=\frac{x_1}{2}$

Now, $0=v^2-2a_1x_2$

substituting the values,

$\Rightarrow 0=v^2-2a_1\frac{x_1}{2}$

$\Rightarrow a_1=\frac{v^2}{x_1}$

$\Rightarrow \mu_k g=\frac{2g\sin\theta x_1}{x_1}$

$\Rightarrow \mu_k = 2\sin\theta$