Answer to Question #90611 in Mechanics | Relativity for Austin

Question

An amusement ride performs 1750 J of work moving a rider over a distance of 4.5 meters. What was the force acting in the rider?

Answer 1

Assuming that the force is directed along the ride direction and is constant, the following expression can be used:

$$A =Fs$$

Hence, the magnitude of the force is

$$F = \frac{A}{s}$$

Substituting the numerical values, we obtain:

$$F = \frac{1750}{4.5} \approx 389 \, N$$

Answer: 389 N.