Question

A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 28.4 kg and the mass of the sled is 15.3 kg.

The force of friction acting on the sled as it slides is -96 N.

If the sled has an initial velocity of 41.8 m/s, and it slides for 3.76 seconds before it hits a wall.

(Ignoring the unfortunate mess with the wall) How much work did friction do during this time?

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*Hint: you cannot solve the problem in 1 step with the current information, you will need to find acceleration, and use time to find something else before you can find an answer!

Answer 1

Answer:

-13,594 J

Explanation:

First of all, we have to find the acceleration of the girl+sled system. This can be found by using Newton's second law of motion:

$F=ma$

where:

F = -96 N is the net force (the force of friction) acting on the system

m = 28.4+15.3 = 43.7 kg is the total mass of the girl and the sled

a is their acceleration

Solving for a,

$a=\frac{F}{m}=\frac{-96}{43.7}=-2.2 m/s^2$

The negative sign means the sled is slowing down.

Now we can find the displacement of the sled during the deceleration phase, by using the suvat equation:

$s=ut+\frac{1}{2}at^2$

where

u = 41.8 m/s is the initial velocity

t = 3.76 s is the time

$a=-2.2 m/s^2$ is the acceleration

So,

$s=(41.8)(3.76)+\frac{1}{2}(-2.2)(3.76)^2=141.6 m$

Now we can find the work done by friction on the sled, which is given by:

$W=Fs$

where

F = -96 N is the force of friction

s = 141.6 m is the displacement of the sled+girl system

Solving,

$W=(-96)(141.6)=-13,594 J$

where the negative sign means the force is opposite to the displacement.