## tacotime one year ago 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?

1. Michele_Laino

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 $$d=1.2 \: m$$

2. Michele_Laino

of course, $$m$$ is the mass of your box

3. tacotime

thank you so much