Question

On a frozen pond, a 11 kg sled is given a kick that imparts to it an initial speed of 4.2 m/s. The coefficient of kinetic friction between sled and ice is 0.12. Find the distance the sled moves before coming to rest. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.

Answer 1

Answer:

Distance: 7,5m

Explanation:

To find the frictional force that tends to stop the sled we have to multiply the coefficient of friction to the the Normal force between the sled and the pond surface:

$F_{fr}=C_{fr}*N$

Inthis case this Normal force is just the weight of the sled, obtained multiplying the mass and the gravitational acceleration:

$N=11kg*9,8m/s^{2}=107,8N$

So:

$F_{fr}=0,12*107,8N=12,94N$

This force imparts an acceleration on the sled in the opossite direction of the movement(deceleration) given by Newton´s second law:

$a=\frac{F_{fr} }{m}=\frac{12,94N}{11kg}=1,176m/s^{2}$

The velocity at any time can be calculated as:

$V=V_{0}-a*t$

It comes to rest when V=0, solving for t:

$V=V_{0}-a*t=0 => t=V_{0}/a=(4.2m/s)/(1,176m/s^{2})=3,57s$

To find the distance at this time:

$X= V_{0}*t-1/2*a*t^{2}$

Replacing for t above:

$X= 4.2m/s*3,57s-1/2*1,176m/s^{2}*(3,57s)^{2}=7,5m$