Answer to Question #90613 in Mechanics | Relativity for Mary

Question

A 56-kg rider on a slingshot ride gets launched from the rest to a final velocity of 13 m/s in a time of 1.4 seconds. What is the force that acts on the rider?

Answer 1

Solution. Change in momentum is proportional to the net external force and the time over which a net force acts. Hence

$$\Delta p = F\Delta t .$$

Therefore

$$F=\frac {\Delta p} {\Delta t}.$$

From the definition of the linear momentum get

$$\Delta p = m(v_2 - v_1).$$

As result

$$F=\frac { m(v_2 - v_1)} {\Delta t}.$$

According to the condition of the problem m=56kg, v2=13 m/s, v1=0 m/s, change in time t=1.4s. Hence

$$F=\frac { 56(13 -0)} {1.4}=520N.$$

Answer. 520 N.