Question

What average net force is required to stop an 8500 kg truck in 10 s if it’s initially traveling at 20 m/s?

Answer 1

We have: F = m × v/t
Here, m = 8500 Kg
v = 20 m/s
t = 10 s

Substitute their values into the expression, 
F = 8500 × 20/10
F = 8500 × 2
F = 17000 N

In short, final answer would be 17000 N

Hope this helps!!