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:

As first thing, we find the acceleration of the system consisting of sled+girl. We do it by using Newton's Second Law of motion:

$F=ma$

where:

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

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

a is the acceleration

Solving for a, we find:

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

Where the negative sign means the direction of the acceleration is opposite to the direction of motion.

Now we find the displacement of the sled, using the suvat equation:

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

where:

u = 41.8 m/s is the initial velocity of the sled

t = 3.76 s is the time elapsed

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

Substituting, we find:

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

Finally, we find the work done by friction on the sled:

$W=Fs$

where:

F = -96 N is the force of friction

s = 141.6 m is the displacement of the system

Substituting:

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

where the negative sign means the force of friction is opposite to the direction of motion.