Question

description: show all you work, including units, on separate paper. follow the \problem solving method\. this final exam is cumulative and covers material from the entire course. instructions: from the list of choices, select the one best answer. multiple attempts: not allowed. this test can only be taken once. force completion: this test can be saved and resumed later. your answers are saved automatically. question completion status: question numbers 1 - 62 shown in a grid question 34 of 62 1.6 points save answer a 75 kg skier coasts 200 m along level snow to a stop from a speed of 14.0 m/s. using the work - energy theorem, what is the coefficient of friction between the skis and the snow? see example 6 - 1 from lecture. options: 0.030, 0.025, 0.075, 0.050

Explanation:

Step1: Recall Work - Energy Theorem

The work - energy theorem states that the net work done on an object is equal to the change in its kinetic energy, \(W_{net}=\Delta KE\). The initial kinetic energy is \(KE_i = \frac{1}{2}mv_i^2\) and the final kinetic energy is \(KE_f=\frac{1}{2}mv_f^2\). Since the skier comes to a stop, \(v_f = 0\), so \(\Delta KE=KE_f - KE_i=-\frac{1}{2}mv_i^2\).

The work done by friction is \(W_f=-f_kd\), where \(f_k=\mu_kN\) and on a level surface \(N = mg\) (normal force equals weight), so \(f_k=\mu_kmg\) and \(W_f=-\mu_kmgd\).

Step2: Set Up Equation from Work - Energy Theorem

From \(W_{net}=\Delta KE\), and the only work done is by friction, we have:
\(-\mu_kmgd=-\frac{1}{2}mv_i^2\)
We can cancel out the mass \(m\) from both sides of the equation:
\(-\mu_kgd=-\frac{1}{2}v_i^2\)
Then, solve for \(\mu_k\):
\(\mu_k=\frac{v_i^2}{2gd}\)

Step3: Substitute Values

We know that \(v_i = 14.0\ m/s\), \(g = 9.8\ m/s^2\), and \(d=200\ m\).
Substitute these values into the formula for \(\mu_k\):
\(\mu_k=\frac{(14.0)^2}{2\times9.8\times200}\)
First, calculate the numerator: \((14.0)^2 = 196\)
Then, calculate the denominator: \(2\times9.8\times200= 3920\)
Now, divide the numerator by the denominator: \(\mu_k=\frac{196}{3920}=0.050\)

Answer:

0.050