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

Distance: 7,5m

Step-by-step 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=(F_(fr) )/(m)=(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`