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 \(p = mv\). Here, \(m = 60\space kg\) and \(v=3\space m/s\). So, \(p_{initial}=60\times3 = 180\space kg\cdot m/s\). The final momentum \(p_{final}=0\) (since the student stops).

Step2: Use the impulse - momentum theorem

The impulse - momentum theorem is \(F_{avg}\Delta t=\Delta p=p_{final}-p_{initial}\). We know \(\Delta t = 0.2\space s\), \(p_{final}-p_{initial}=- 180\space kg\cdot m/s\). Rearranging for \(F_{avg}\), we get \(F_{avg}=\frac{\Delta p}{\Delta t}\).
Substituting the values: \(F_{avg}=\frac{-180}{0.2}=-900\space N\). But we need to consider the force due to gravity. The force due to gravity \(F_g = mg=60\times9.8 = 588\space N\). The net force on the student is the normal force \(N\) (the force the floor exerts on the student). Using \(F_{avg}=N - mg\) (taking upward as positive). Since \(F_{avg} = 900\space N\) (magnitude), then \(N=F_{avg}+mg\).
\(N = 900+588=1488\space N\)

Answer:

The average force on the student is \(1488\space N\)