# A man is pushing a box from behind with a constant net force F=146 N

1. May 6, 2014

### mrf2dhimself

A man is pushing a box from behind with a constant net force F=146 N parallel to the ground on the level floor, moving it to a distance of 54 m. The coefficient of the kinetic friction between the box and the floor is 0.060. Mass of the box is 92 kg. Initially the box was at rest.

a) Calculate the acceleration of the box
b) Calculate its final velocity
c) Calculate the work of the force F
d) Calculate the work of the friction force

2. May 7, 2014

### Astronuc

Staff Emeritus
Please write the equations and show one's effort to solve these problems.

What is the mathematical relationship between force, mass and acceleration?

3. May 7, 2014

### whdahl

You need to find the net force acting on the box which is the force the man is applying minus the friction force (Which always opposes motion). The formula for frictional force is F = μkN where μk is the stated coefficient of friction and N is the weight of the box in Newtons. Now that you have the total force applied on the box, you can use a = F/m to find the acceleration of the box.

b) There are a couple ways to go about finding the final velocity. The first way involves using the equation X = X0 + V0t + 0.5*a*t2 to solve for the total time (t) that it takes for the box to travel X distance (54m). Given the initial condition that the box is at rest, V0 is 0, and we can arbitrarily take X0 to be 0 as well. Now that you have the time t, the final velocity is simply Vf = a*t.

The more simple way requires you to have remembered (or looked up) the kinematic equation Vf2 = V02 + 2*a*X to find the final velocity in one or two steps.

c and d) work is simply Force times distance.