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 student’s hand? (assume initial velocity is zero)

Explanation:

Step1: Calculate the gravitational force

The gravitational force \(F_g = mg\), where \(m = 0.500\space kg\) and \(g=9.8\space m/s^2\).
\(F_g=0.500\times9.8 = 4.90\space N\)

Step2: Calculate the average net force

The average applied force \(F_{app}=8.20\space N\). The net force \(F_{net}=F_{app}-F_g\)
\(F_{net}=8.20 - 4.90=3.30\space N\)

Step3: Use the work - energy theorem for part b

The work done by the net force \(W = F_{net}d\), where \(d = 1.50\space m\). So \(W=3.30\times1.50 = 4.95\space J\)
According to the work - energy theorem \(W=\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mu^2\). Since \(u = 0\space m/s\), we have \(W=\frac{1}{2}mv^2\)
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. \(3.30\space N\)
b. \(4.45\space m/s\)