An ice skater of mass m is given a shove on a frozen pond. After the shove she has a speed of Vo = 2 m/s. Assuming that the only horizontal force...
An ice skater of mass m is given a shove on a frozen pond. After the shove she has a speed of Vo = 2 m/s. Assuming that the only horizontal force that acts on her is a slight frictional force between the blades of the skates and the ice.
(a) Draw a free body diagram.
(b) Find the distance the skater moves before coming to rest if the coefficient of kinetic friction between the blades of the skates and ice is µk = 0.12.
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A.
The free body diagram needs to show all the FORCES acting on the skater. It will not show her mass, velocity or acceleration by themselves.
There are three forces acting on the skater after she is shoved:
 The force of gravity, pulling her down
 The normal force, the electromagnetic repulsion between her atoms and the ground, pushing her up
 The "negative" force, acting behind her, that is slowing her down.
So, the free body diagram should look like this;
Normal Force (mg)
/\

Friction (µkmg)<Skater

\/
Force of Gravity (mg)
We know that the Normal and Gravity forces must be equal, because she's not being pushed off into the air, or sinking into the ice; the forces are equal and opposite.
B.
We can use the WorkEnergy theorem to demonstrate that the skater possesses energy, which is being taken away by friction.
The WorkEnergy Theorem states that a change in energy equals the work done, and vice versa.
We know that Work = Force acting over a distance
W = Fd
We also know that Energy, measured in Joules, has the same units as Work. This means that Work and Energy are interchangeable.
We know the skater starts at a speed of 2m/s, and her final speed will be 0. All of her energy will be lost because of friction. Therefore the work done on her must exactly equal her initial energy.
`KE_i = Fd`
This means that the skater's initial Kinetic Energy must equal the Force acting on her, times the distance it takes her to stop.
What is the Force acting on her? Friction.
What is Friction? µk(mg)
We also know that `KE = .5mv^2`
So to rearrange the equation, we get
.5mv^2 = µk(mg)d
The masses cancel out, leaving us with
.5v^2 = µk(g)d
We know V, µk and g:
.5(2^2) = .12(9.8)d
Solving for d, we get
`(.5(4))/(.12(9.8))` = d
d = 1.7m