Question

1. a 0.500 kg object is thrown vertically upward with an average applied force of 8.20 n by a student. the force is applied through a displacement of 1.50 m.

a. what is the average net force acting on the object?
b. what is the velocity of the object when it leaves the students hand? (assume initially v = 0)

Explanation:

Step1: Calculate the force due to gravity

The force due to gravity \(F_g = mg\), where \(m = 0.500\space kg\) and \(g=9.8\space m/s^{2}\).
\(F_g=0.500\times9.8 = 4.9\space N\)

Step2: Find the net force

The net force \(F_{net}=F_{applied}-F_g\). Given \(F_{applied} = 8.20\space N\)
\(F_{net}=8.20 - 4.9=3.3\space N\)

Step3: Use work - energy theorem for part b

The work done by the net force \(W = F_{net}d\), where \(d = 1.50\space m\). So \(W=3.3\times1.50 = 4.95\space J\)
According to the work - energy theorem \(W=\Delta K.E=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}\). Since \(u = 0\) (initial velocity), \(W=\frac{1}{2}mv^{2}\)
We can solve for \(v\): \(v=\sqrt{\frac{2W}{m}}\)
Substitute \(W = 4.95\space J\) and \(m = 0.500\space kg\)
\(v=\sqrt{\frac{2\times4.95}{0.500}}=\sqrt{19.8}\approx 4.45\space m/s\)

Answer:

a. The average net force is \(3.3\space N\)
b. The velocity of the object when it leaves the student's hand is approximately \(4.45\space m/s\)