Question

A skater pushes on the back of a 45.3 kg sled with an average force of 72.5 N over a distance of 15.6 m. Find the final speed of the sled if it was moving at 1.31 m/s initially. (Use the work-energy theorem)

Answer 1

Answer:

7.19 m/s

Explanation:

By the work-energy theorem, we have the following equation

$\begin{gathered} W=K_f-K_i \\ Fd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{gathered}$

Where F is the force, d is the distance, m is the mass, vf is the final speed and vi is the initial speed. Solving for vf, we get

$\begin{gathered} Fd+\frac{1}{2}mv_i^2=\frac{1}{2}mv_f^2 \\ 2(Fd+\frac{1}{2}mv_i^2)=mv_f^2 \\ \\ \frac{2(Fd+\frac{1}{2}mv_i^2)}{m}=v_f^2 \\ \\ v_f=\sqrt{\frac{2(Fd+\frac{1}{2}mv_i^2)}{m}} \end{gathered}$

Replacing F = 72.5 N, d = 15.6 m, m = 45.3 kg, and vi = 1.31 m/s, we get

$\begin{gathered} v_f=\sqrt{\frac{2(72.5N(15.6m)+\frac{1}{2}(45.3kg)(1.31\text{ m/s\rparen}^2)}{45.3\text{ kg}}} \\ \\ v_f=\sqrt{\frac{2(1131\text{ N m + 38.87 N m\rparen}}{45.3\text{ kg}}} \\ \\ v_f=\sqrt{\frac{2339.74\text{ N m}}{45.3\text{ kg}}} \\ \\ v_f=\sqrt{51.65\text{ m}^2\text{ /s}^2} \\ v_f=7.19\text{ m/s} \end{gathered}$

Therefore, the final speed of the sled was 7.19 m/s