Kinetics Question

Question

When an ice hockey puck weighing 90 g is stationary on the ice, a minimal force of 0.2 N is required to move it.

a) Calculate the coefficient of static friction.

b) If the puck covers the first 10 cm in 0.5 s (while the 0.2 N force is still applied on the puck), calculate the kinetic friction coefficient.

Answer 1

First, draw a free body diagram:

You should have been instructed that the coefficient of friction (μ) is used to find the frictional force like this:

F_Friction = μ•F_Normal or, in this case, μ = F_Friction/F_Normal

With regards the the STATIC coefficient of friction, that is the coefficient for which the force just breaks the object loose from the surface, which is 0.2 N.

To find the normal force, consider that there is no motion or acceleration in the y-direction. That means the sum of the forces in the y-direction equals zero because F = ma and "a" = zero.

∑F_y = F_Normal - W where W is the weight (which is the mass times the acceleration due to gravity) so

∑F_y = 0 = F_Normal - mg

F_Normal = mg

F_Normal = (0.09 kg)(9.8 m/s²) = 0.882 N

In the x-direction, since the Applied force is pushing just enough to break the object free, that just momentarily prior to it breaking loose, there is no motion so ∑F_x = 0 as well so:

∑F_x = 0 = F_Applied - F_Friction

F_Applied = F_Friction

0.2N = F_Friction

Using μ = F_Friction/F_Normal :

μ_static = 0.2/0.882 = 0.2

To find the kinetic coefficient of friction, find the acceleration of the object using the kinematic equations of motion (assuming constant acceleration): x = x₀ + v_ot + 1/2at² x₀ = 0 and v₀ = 0 so x = 1/2at² or a = 2x/t² = 2(0.1 m)/(0.5)² = 0.8 m/s²

The ∑F_x = ma = (0.09)(0.8) = 0.072 N

The applied force is 0.2N so the frictional force must be 0.2 - 0.072 = 0.128

μ = F_Friction/F_Normal :

μ_kinetic = 0.128/0.882 = 0.1

Answer 2

The minimal force required to move it is going to be the minimum force required to match(or just barely exceed) the force of static friction keeping it in place. This minimum force of 0.2N is thus equal to N*mu(Force normal times the coefficient of friction). The normal force will be equally balanced by the force of gravity in the y direction, which is given as mass*gravitational acceleration = 0.09kg*9.81m/s^2 = 0.8829N, which is the magnitude of both the gravitational force and the normal force.

Thus the force required to move the puck is equal to the coefficient of friction times the normal force:

0.2N = 0.8829N * mu

We can solve this for the coefficient of friction, mu, by dividing through by 0.8829N, and we get mu = 0.2265.

Now to calculate the kinetic friction in the second half of the problem, we need to know the net force the puck feels. We can obtain this through the acceleration value, because net fore = mass*acceleration by Newton's second law.

To find the acceleration, we will use the 2nd kinematic: delta x = initial velocity*time + 0.5* acceleration*time^2

delta x is 0.1 m

initial velocity is 0

time is 0.5 s

Thus 0.1 = 0*0.5 + 0.5*a*0.5^2, and solving for a, we get 0.8 m/s^2

Then our net force, by newton's second law is 0.09kg*0.8m/s^2 = 0.072N

If our applied force is 0.2N, and our resulting force is 0.072N, then the resistive force is 0.2-0.072= 0.128N

Now this is the kinetic friction force, which is again equal to the normal force(0.8829N, the y direction hasnt changed), times the coefficient of kinetic friction, mu.

0.128N = 0.8829N*mu, and solving again for mu, we get 0.145.

Answer 3

a) Force of static friction = μ_s x Normal force, where μ_s is the coefficient of static friction. So if it takes 0.2 N just to move it, 0.2 = μ_s x mg = μ_s x 0.09 kg x 9.8 m/s^2.

μ_s = 0.2N / 0.09 kg x 9.8 m/s^2. = 0.227.

b) When it is moving, the kinetic frictional force opposes the applied force, so it is the net force that causes the motion F_net = 0.2 N - μ_k x mg.

If the puck covers 0.1 m in 0.5 s, its acceleration can be obtained from D = V₀ t + 1/2 x a x t².

As V₀ is zero, this becomes D = 1/2 x a x 0.5² and a = 2 x 0.1 m / 0.5² = 0.8 m/s².

The net force is then ma = 0.09 kg x 0.8 m/s² = 0.072 N and this is equal to 0.2 N - μ_k x 0.09 kg x 9.8 m/s².

Solving for μ_k = 0.072 N - 0.2 N / -0.882 N = 0.145.

I didn't round to the number of sig figs.