Question

1. a 125 kg runner is traveling at an acceleration of 2.1 m/s². if friction applies 40 n of force, what is the force applied by the runner?

250 n
290 n
287 n
210 n

Explanation:

Step1: Apply Newton's second law

According to Newton's second law \(F = ma\). Here, \(m = 125\space kg\) and \(a=2.1\space m/s^{2}\). So the net force \(F_{net}=ma = 125\times2.1\space N\).

$$F_{net}=125\times2.1=262.5\space N$$

Step2: Relate applied force and net force

The net force \(F_{net}\) is given by \(F_{applied}-F_{friction}\). We know \(F_{friction} = 40\space N\) and \(F_{net}=262.5\space N\). Then \(F_{applied}=F_{net}+F_{friction}\)

$$F_{applied}=262.5 + 40=302.5\approx303\space N$$

Answer:

303 N