Question

1. a 125 kg runner is traveling at an acceleration of 2.8 m/s². if friction applies 40 n of force, what is the force applied by the runner? 150 n 390 n 197 n 310 n

Explanation:

Step1: Apply Newton's second law

Newton's second law is \(F_{net}=ma\). Here, \(m = 125\space kg\) and \(a=2.8\space m/s^{2}\). So, \(F_{net}=125\times2.8\).

$$F_{net}=125\times2.8 = 350\space N$$

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

The net force \(F_{net}\) is the difference between the force applied by the runner \(F_{applied}\) and the frictional force \(F_{friction}\). That is \(F_{net}=F_{applied}-F_{friction}\). We know \(F_{net} = 350\space N\) and \(F_{friction}=40\space N\). Then \(F_{applied}=F_{net}+F_{friction}\).

$$F_{applied}=350 + 40=390\space N$$

Answer:

390 N