Question

A 4.30 kg box is pushed 1.20 m with a 36 N force. What is the final speed of the box if the coefficient of kinetic friction is 0.30?

Answer 1

on your box are acting two forces, namely:
the driving force F=20 N
and the friction force: R=\mu*m*g=0.3*4.30*9.81=...N
Now the work done by those forces has to be equal to the kinetic energy change, so we can write this equation:
\[Fd - Rd = \frac{1}{2}m{v^2}\]
where \(v\) is the requested final speed, and \(d\) is the distance traveled by your box, namely \(d=1.2 \: m\)

Answer 2

of course, \(m\) is the mass of your box

Answer 3

thank you so much