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.

Answer 1

7.19 meters per second

Step-by-step explanation

Step 1

find the acceleration of the skater

Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate,it is defined by the expression

`\begin{gathered} F=ma \\ where\text{ F is the applied force} \\ m\text{ is the mass} \\ a\text{ is the accelearaion} \end{gathered}`

so

a) Let

`\begin{gathered} F=72.5\text{ N} \\ m=45.3\text{ Kg} \end{gathered}`

b) now, replace and solve for a

`\begin{gathered} F=ma \\ replace \\ 72.5\text{ N=45.3 kg*a} \\ divide\text{ both sides by 45.3 Kg} \\ (72.5N)/(45.3)\text{=a} \\ a=1.6\text{ }(m)/(s^2) \end{gathered}`

Step 2

Now, to find teh final speed we need to use the expression

`v_f^2=v_1^2+2ax`

so

a)Let

`\begin{gathered} v_i=1.31(m)/(s) \\ x=15.6m \\ a=1.6(m)/(s^2) \end{gathered}`

replace

`\begin{gathered} v_f^2=v_1^2+2ax \\ v_f^2=(1.31)^2+2(1.6)(15.6) \\ v_f^2=51.63 \\ v_f=√(51.63) \\ v_f=7.19(m)/(s^) \end{gathered}`

so, the answer is

7.19 meters per second

I hope this helps you