Question #90611
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?
1
Expert's answer
2019-06-07T11:52:56-0400

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


A=FsA =Fs

Hence, the magnitude of the force is


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

Substituting the numerical values, we obtain:


F=17504.5389NF = \frac{1750}{4.5} \approx 389 \, N

Answer: 389 N.


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