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

7.19 m/s

Step-by-step explanation:

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

`\begin{gathered} W=K_f-K_i \\ Fd=(1)/(2)mv_f^2-(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+(1)/(2)mv_i^2=(1)/(2)mv_f^2 \\ 2(Fd+(1)/(2)mv_i^2)=mv_f^2 \\ \\ (2(Fd+(1)/(2)mv_i^2))/(m)=v_f^2 \\ \\ v_f=\sqrt{(2(Fd+(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)+(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