Question

1. an 85 kg bicyclist is traveling at an acceleration of 3.7 m/s². if friction applies 25 n of force, what is the force applied by the bicyclist? 350 n 177.5 n 289.5 n 339.5 n

Explanation:

Step1: Use Newton's second - law formula

According to Newton's second - law $F = ma$, where $m = 85$ kg and $a=3.7$ m/s². First, find the net force $F_{net}=ma$.
$F_{net}=85\times3.7$
$F_{net} = 314.5$ N

Step2: Relate net - force, frictional force and applied force

The net force $F_{net}$ is the difference between the force applied by the bicyclist $F_{applied}$ and the frictional force $F_{friction}$. That is $F_{net}=F_{applied}-F_{friction}$.
We know $F_{net} = 314.5$ N and $F_{friction}=25$ N.
So, $F_{applied}=F_{net}+F_{friction}$
$F_{applied}=314.5 + 25$
$F_{applied}=339.5$ N

Answer:

339.5 N