Question

A small car of mass 643 kg is parked behind a small truck of mass 1415 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 43 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 2 m/s². What is the acceleration of the truck? Answer in units of m/s².

Answer 1

Answer:

0.882 m/s²

Explanation:

m = Mass of car = 643 kg

a = Acceleration of the car = 2 m/s²

$a_2$ = Acceleration of the truck

Force exerted on the car

$F=ma\\\Rightarrow F=643\times 2\\\Rightarrow F=1286\ N$

Mass of the truck with the woman = $M=1415+43=1458\ kg$

From Newton's second law

$F=Ma_2\\\Rightarrow a_2=\frac{1286}{1458}\\\Rightarrow a=0.882\ m/s^2$

The acceleration of the truck is 0.882 m/s²