Question

Two people are competing in a tubing competition. They push a 132 kg inner

tube. If the combined push of the team is 450 N and the inner tube also
experiences a friction force of 35 N, what is the acceleration of the inner tube? (F
=ma)

Answer 1

`a=3.14\ m/s^2`

Step-by-step explanation:

Given that,

Mass of the tube, m = 132 kg

Force = 450 N

Friction force, f = 35 N

Net force,

F = 450 - 35

F = 415 N

Let a be the acceleration. We know that,

F = ma

`a=(F)/(m)\\\\a=(415)/(132)\\\\a=3.14\ m/s^2`

So, the acceleration is
`3.14\ m/s^2`.