Question

A curling stone of mass \(19.00 \mathrm{~kg}\) is released with an initial speed \(v_{0}=3.070 \mathrm{~m} / \mathrm{s}\) and slides on level ice. The curling stone travels \(36.21 \mathrm{~m}\) before it stops. What is the coefficient of kinetic friction between the curling stone and the ice?

Short answer

The coefficient of kinetic friction between the curling stone and the ice is approximately \(0.01635\).

Step-by-step solution

Write down the information given

We have the following information: Mass of the curling stone (\(m\)): \(19.00 \mathrm{~kg}\) Initial speed (\(v_0\)): \(3.070 \mathrm{~m} / \mathrm{s}\) Distance traveled before stopping (\(d\)): \(36.21 \mathrm{~m}\)

Write down the work-energy theorem

The work-energy theorem states: \(W = \Delta KE\) Where \(W\) is the work done by the force, and \(\Delta KE\) is the change in kinetic energy.

Calculate the change in kinetic energy

Since the curling stone comes to a stop, its final kinetic energy (\(KE_f\)) will be zero. The initial kinetic energy (\(KE_i\)) is given by: \(KE_i = \frac{1}{2}mv_0^2\) So the change in kinetic energy is: \(\Delta KE = KE_f - KE_i = -\frac{1}{2}mv_0^2\)

Calculate the work done by the friction force

The work done by the friction force is given by: \(W = F_fd\) Where \(F_f\) is the friction force and \(d\) is the distance the object traveled.

Express the friction force in terms of the coefficient of kinetic friction

The friction force is given by: \(F_f = \mu_k F_N\) Where \(\mu_k\) is the coefficient of kinetic friction and \(F_N\) is the normal force. Since the curling stone is on level ice (horizontal plane), the normal force is equal to the weight of the stone, which is \(mg\). Hence, \(F_f = \mu_k mg\)

Substitute the expressions of work and friction force into the work-energy theorem

Using the previously derived expressions for work and friction force, we can write the work-energy theorem as: \(\mu_k m gd = -\frac{1}{2}mv_0^2\)

Solve for the coefficient of kinetic friction

Divide both sides by \(mgd\) and solve for \(\mu_k\): \(\mu_k = \frac{-\frac{1}{2}mv_0^2}{mgd}\) Plug in the given values (\(m=19.00 \mathrm{~kg}\), \(v_0=3.070 \mathrm{~m} / \mathrm{s}\), \(d=36.21 \mathrm{~m}\), and \(g=9.81 \mathrm{~m} / \mathrm{s}^2\)): \(\mu_k = \frac{-\frac{1}{2}(19.00)(3.070)^2}{(19.00)(9.81)(36.21)}\) Calculate the value of \(\mu_k\): \(\mu_k \approx 0.01635\) So, the coefficient of kinetic friction between the curling stone and the ice is approximately \(0.01635\).