Question

Consider mechanical energy to find the coefficient of friction between the sledge and the ground

Answer 1

Answer:

0.08

Explanation:

We can represent the situation with the following figure

Now, by the conservation of energy, we can write the following equation

$\begin{gathered} K_i+U_i-Wnc=K_f \\ \frac{1}{2}mv_i^2+mgh-F_fd=\frac{1}{2}mv_f^2 \end{gathered}$

Where m is the mass, vi is the initial velocity, g is the gravity, h is the height, Ff is the force of friction, d is the distance traveled by the sledge, and vf is the final velocity.

Using the free body diagram, we get that the force of friction is equal to

$\begin{gathered} F_n=mg\sin60 \\ \text{ Then} \\ F_f=\mu F_n \\ F_f=\mu mg\sin60 \end{gathered}$

Now, we can replace the expression for Ff in the equation above and solve for the coefficient of friction μ

$\begin{gathered} \frac{1}{2}mv_i^2+mgh-(\mu mg\sin60)d=\frac{1}{2}mv_f^2 \\ \\ \frac{1}{2}v_i^2+gh-\mu gd\sin60=\frac{1}{2}v_f^2 \\ \\ \mu gd\sin60=\frac{1}{2}v_i^2+gh-\frac{1}{2}v_f^2 \\ \\ \mu=\frac{1}{gd\sin60}(\frac{1}{2}v_i^2+gh-\frac{1}{2}v_f^2) \\ \end{gathered}$

Replacing g = 9.8 m/s², d = 16 m, vi = 3 m/s, h = 8 m, and vf = 12 m/s, we get

$\begin{gathered} \mu=\frac{1}{(9.8)(16)\sin60}(\frac{1}{2}(3)^2+(9.8)(8)-\frac{1}{2}(12)^2) \\ \\ \mu=0.08 \end{gathered}$

Therefore, the coefficient of friction is 0.08