Question

1. what net force is required to stop a 10,000 kg truck in 50 seconds if its initially traveling at 20 m/s?

Explanation:

Step1: Find acceleration

Acceleration \(a\) is change in velocity over time. Final velocity \(v = 0\) (stops), initial velocity \(u = 20\ m/s\), time \(t = 50\ s\).
\(a=\frac{v - u}{t}=\frac{0 - 20}{50}=-\frac{20}{50}=-0.4\ m/s^{2}\)

Step2: Calculate net force

Using Newton's second law \(F = ma\), mass \(m = 10000\ kg\), \(a=-0.4\ m/s^{2}\).
\(F = 10000\times(-0.4)= - 4000\ N\) (negative sign indicates force opposite to motion direction, magnitude is 4000 N).

Answer:

The net force required is \(\boldsymbol{4000\ N}\) (magnitude, direction opposite to motion).