Question

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N pushes back against the motion.

What is the acceleration of the cart?

Answer 1

0.8214 m/s^2

Step-by-step explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2