Question

A girl coasts down a hill on a sled, reaching level ground at the bottom with a speed of 7.1 m/s. The coefficient of kinetic friction between the sled runners and the hard, icy snow is 0.045, and the girl and sled together weigh 783 N. The acceleration of gravity is 9.81 m/s 2 . How far does the sled travel on the level ground before coming to a rest?

Answer 1

Answer:

114.19186 m

Explanation:

v = Velocity of girl at bottom = 7.1 m/s

$\mu$ = Coefficient of kinetic friction = 0.045

g = Acceleration due to gravity = 9.81 m/s²

d = Distance

N = Normal force = 783 N

The weight of the sled and girl is considered as the sum of the frictional force and weight

Hence we use the following equation where the kinetic energy and potential energy are conserved

$\frac{1}{2}mv^2=\mu Nd\\\Rightarrow \frac{1}{2}\frac{783}{9.81}\times 7.1^2=0.045\times 783\times d\\\Rightarrow d=\frac{1}{2}\times \frac{783\times 7.1^2}{9.81\times 0.045\times 783}\\\Rightarrow d=114.19186\ m$

The sled travels 114.19186 m on the level ground before coming to a rest