Question

A net force of 25N is applied for 5.7s to a 12kg box initially at rest. What is the speed of the box at the end of the 5.7s interval
a)12m/s
b)1.8m/s
c)30m/s
d)3.0m/s
e)7.5m/s
Please explain as I need help on learning how to do this

Answer 1

You will need Newton Second law \[F =ma\]And you will need a kinematic equation to get final velocity after acceleration has been applied. We are going to assume acceleration does not vary so acceleration is constant. So we can use this equation\[Vf = Vi + at\]

We know F/m = a and t is time in seconds.
Vf = Vi + at.

\[V_{f} = V _{i} + at\] But since a =F/m we can do this \[V_{f}=V _{i} + (\frac{ F }{ m }) t\]
You know initial velocity is v=0 then the equation becomes \[V _{f}= (\frac{ F }{ m})t\]

Answer 2

From here its plugging in the values you are given.

Answer 3

so I use the last equation correct? @snowsurf

Answer 4

Yes