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)

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].