An ice skater of mass m is given a shove on a frozen pond. After the shove she has a speed of 2m/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.) Identify the horizontal force and two vertical forces that act on her.
b.) Use the work-energy theorem to find the distance the skater moves before coming to rest. Assume that the coefficient of kinetic friction between the blades of the skates and ice is .12
horizontal friction
vertical gravity down, force up from ice equal and opposite
initial Ke = (1/2) m v^2 = .5*4 m = 2 m Jloules
final Ke = 0
normal force at ice = 9.81 * m
so
friction force = .12 (9.81) m
work done = Force * distance
= .12 * 9.81* m * d = change in ke = 2 m
so
d = .12*9.81/2 meters
d=2/(.12*9.81)
a.) The horizontal force acting on the skater is the frictional force between the blades of the skates and the ice. This force opposes the motion of the skater and is responsible for slowing her down.
The two vertical forces acting on the skater are her weight, which acts downwards and is equal to mg (where g is the acceleration due to gravity), and the normal force exerted by the ice on the skater, which acts upwards and balances her weight.
b.) The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the frictional force will be equal to the change in the skater's kinetic energy.
The initial kinetic energy of the skater is given by (1/2)mv^2, where m is the mass of the skater and v is her initial velocity. The final kinetic energy is zero since she comes to rest.
The work done by the frictional force is equal to the force of friction multiplied by the distance over which it acts. The force of friction can be calculated using the equation F = μN, where μ is the coefficient of kinetic friction and N is the normal force.
Since the normal force and the weight of the skater are equal in magnitude but act in opposite directions, the normal force can be calculated as N = mg.
Therefore, the work done by the frictional force is W = μmgd, where d is the distance the skater moves before coming to rest.
Setting the work done equal to the change in kinetic energy, we have:
W = ΔKE
μmgd = (1/2)mv^2 - 0
μmgd = (1/2)mv^2
d = (1/2)(v^2)/(μg)
Substituting the given values μ = 0.12, v = 2 m/s, and g ≈ 9.8 m/s^2, we can calculate the distance the skater moves before coming to rest:
d = (1/2)(2^2)/(0.12*9.8)
d = 1/2 * 4/1.176
d ≈ 3.40 meters
Therefore, the skater moves approximately 3.40 meters before coming to rest.