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)
Question
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:
[tex]a=3.14\ m/s^2[/tex]
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
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{415}{132}\\\\a=3.14\ m/s^2[/tex]
So, the acceleration is [tex]3.14\ m/s^2[/tex].