Question

A girl coasts down a hill on a sled, reaching

Answer 1

The sled travels a distance of 60.2 m before coming to rest.

The girl  reaches the ground with a speed u. She then travels forward on the hard icy snow, which has a coefficient of kinetic friction $\mu _k$ . She comes to rest after travelling a distance s on the ground, due to the friction between the sled and the ground.

If the force of friction is f , the mass of the girl and the sled is m and the acceleration due to gravity is g, then,

$f=\mu_kmg$    .......(1)

This force exerts a decelerating force on the sled. If the deceleration of the sled is a, then,

$f=ma$   .......(2)

From equations (1) and (2),

$f=\mu_kmg=-ma\\ a=-\mu_kg$......(3)

Substitute 0.038 for  $\mu _k$ and 9.81 m/s²for g.

$a=-\mu_kg\\ =-(0.038)(9.81m/s^2)\\ =-0.3728m/s^2$

Since the girl comes to rest, its final velocity is 0. Substitute 6.7 m/s for u and -0.3728m/s² for a in the equation of motion

$v^2=u^2+2as$

Solve for s.

$v^2=u^2+2as\\ (0m/s)^2=(6.7m/s)^2+2(-0.3728m/s^2)s\\ s=\frac{(6.7m/s)^2}{2(0.3728m/s^2)} \\ =60.2m$

Thus, the girl travels a distance of 60.2 m before coming to rest.