A 3.6 kg object is accelerated from rest to a
speed of 22.9 m/s in 34 s.What average force was exerted on the object during this period of acceleration?
Answer in units of N

Question

A 3.6 kg object is accelerated from rest to a
speed of 22.9 m/s in 34 s.What average force was exerted on the  object during this period of acceleration?
Answer in units of N

anonymous · Answer

You can use the equation Vf=Vi+at in order to find the acceleration. Vf is final velocity, or 22.9, and Vi is initial velocity, which is zero, so the equation is Vf=at.
Divide by t in order to solve for acceleration.
a=Vf/t
You are given Vf and t, so plug them in.
a=22.9/34
a=.674 m/s^2
This should be the average acceleration over the 34 seconds.
Now, in order to find force, use Newton's second law, F=ma.
F=ma
F=3.6*.674
F=2.425 N

Hope that helps!

anonymous · Answer

thank you poeticalto