You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor at constant speed. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by you on the crate?

I have determined that the -1986.53J is the resistive force of friction, but how the hell can i find the work done on the crate? wouldnt it be any number greater than or equal to 1986.53J? but that cant be the answer, because there is no way to express that on this HW. All I want is some help getting to the answer, please dont provide the answer! Thanks guys!

The work is given as F dotted with dl (the distance, you might say). Integration techniques can be used in problems with work, but this one will be fairly simple.
Assuming you know how to work dot products, this one is simple because the angle is 0 degrees, so cos(theta) = 1.
Thus the work in scenarios like this is just the force that you are applying times the distance.

Hopefully this info helps, best I can do without really giving any answer.

I will add that Joules is not a measurement of force. Joules is a unit of energy. Units can be a life saver.

This post was edited by ringo794 on Mar 18 2015 09:05pm

What is the direction of static friction and kinetic friction?

Always in the opposite direction of the motion.

/e Remember, you only need to bust out the integration if the force actually varies as distance changes.

This post was edited by Casey on Mar 19 2015 10:06am

haha, there is another stuff that you have forgotten about inertia.

The wooden crate will be in inertia if it is stopped or with a constant velocity.

well, If the crate has a constant velocity, as you said on this question, then there is no acceleration and the force that is made on it has the same value of the force of friction.

as the body is already in motion, then the friction force is calculated with the kinetic coefficient of friction.

There you go:





Now talking about work

The force F has the same direction of the motion, so it does a positive work. In other way, the Fr is against the motion, so the work is negative.

So, F = Fr

Fr = N. μ (in this case will be 0,3)
Fr = m . g . μ
let's assume g = 9,8 m/s^2
Fr = 150 . 9,8 . 0,3
Fr = - 441 N (of course it is a negative force because is against the motion)

Now let's calculate the work done of the friction:
work = w
distance = d

w = Fr . d
w = -441 . 4,5
w = - 1984,5 J


Now let's see what happening on the crate.
There is a friction doing a work against the motion, ok. Also, in order to keep the crate in motion, it must have another force in the same way of the motion doing some work too, otherwise the crate would be stopped on his position.
The force F is doing the same work of the friction force, since the crate is in inertia.
So, the work of the force F is 1984,5 J.

And the whole work done in the wooden crate is Zero, since the crate is in inertia.
So, there is nothing to calculate on these situations


I hope I could help. Enjoy!