Question

1 a 60 - kilogram student jumps down from a laboratory counter. at the instant he lands on the floor, his speed is 3 meters per second. if the student stops in 0.2 seconds, what is the average force on the student?

Explanation:

Step1: Calculate the initial momentum

Momentum formula is \(p = mv\). Given \(m = 60\space kg\) and \(v=3\space m/s\), so \(p_{initial}=60\times3 = 180\space kg\cdot m/s\).

Step2: Calculate the change in momentum

Final velocity \(v_{final} = 0\space m/s\), so \(p_{final}=60\times0=0\space kg\cdot m/s\). The change in momentum \(\Delta p=p_{final}-p_{initial}=0 - 180=- 180\space kg\cdot m/s\).

Step3: Use the impulse - momentum theorem

The impulse - momentum theorem is \(F_{avg}\Delta t=\Delta p\). We know \(\Delta t = 0.2\space s\) and \(\Delta p=-180\space kg\cdot m/s\). Rearranging for \(F_{avg}\), we get \(F_{avg}=\frac{\Delta p}{\Delta t}\). Substituting values: \(F_{avg}=\frac{- 180}{0.2}=-900\space N\). But we also need to consider the force of gravity \(F_g = mg\) where \(g = 10\space m/s^{2}\), \(F_g=60\times10 = 600\space N\). The net force \(F_{net}=F_{avg}+F_g\). Let the force exerted by the floor be \(F\). Using \(F - mg=\frac{\Delta p}{\Delta t}\), \(F=\frac{\Delta p}{\Delta t}+mg\). Substituting \(\frac{\Delta p}{\Delta t}=- 900\space N\) (magnitude is \(900\space N\)) and \(mg = 600\space N\), \(F=900 + 600=1500\space N\)

Answer:

\(1500\space N\)