Finding coefficient of friction and the mass

1. The problem statement, all variables and given/known data
Ben pushes a box along a floor against a constant force of friction. When Ben pushes with a horizontal force of 75N the acceleration of the box is 0.5 m/s/s; when he increases the force to 81N the acceleration is 0.75m/s/s.


2. Relevant equations

Find the mass of the box and the coefficient of friction between the box and the floor.


3. The attempt at a solution

i used the formula ma = uk*mg and that allows to cancel out mass and get a/g = uk
with that i get .051 but i really dont think that is right im not really sure where the 2 different accelerations and forces come into play...
 

LowlyPion

Homework Helper
3,055
4
1. The problem statement, all variables and given/known data
Ben pushes a box along a floor against a constant force of friction. When Ben pushes with a horizontal force of 75N the acceleration of the box is 0.5 m/s/s; when he increases the force to 81N the acceleration is 0.75m/s/s.


2. Relevant equations

Find the mass of the box and the coefficient of friction between the box and the floor.


3. The attempt at a solution

i used the formula ma = uk*mg and that allows to cancel out mass and get a/g = uk
with that i get .051 but i really dont think that is right im not really sure where the 2 different accelerations and forces come into play...
Welcome to PF.

You are pushing with a constant force. Where does that force go to?

Write an equation for how the 75N is used up.
 
well not really sure what your saying by that but with uk*F = m i can get the mass but im really not sure how to use the 75N to first get uk or the mass.... thats what im missing
 

rock.freak667

Homework Helper
6,232
27
So we know that Fnetmanet.

If the horizontal force is F' and the force of friction is F, then the net force is given by

Fnet=F'-F
so that manet=F'-F

In the question you are given that the horizontal force is 75N and the net acceleration is 0.5ms-2. You can now form one equation. Use the other information given to form the other equation.

You will now have two equations in m and F, solve for the two of them. You can now get the coefficient of friction.
 
yea thanks that is the basic step for the formula with that i was able to get :

m*a = F - Ff
F(75) - Ff = m*a(.5)
F(81) - Ff = m*a(.75)

so from there: (using Ff)

m*a(.5) + F(75) = F(81) + m*a(.75)

that really dosn't seem to fit a lot of the formulas make sence to me but i just feel like im missing something!
 

rock.freak667

Homework Helper
6,232
27
yea thanks that is the basic step for the formula with that i was able to get :

m*a = F - Ff
F(75) - Ff = m*a(.5)....1
F(81) - Ff = m*a(.75)...2

so from there: (using Ff)

m*a(.5) + F(75) = F(81) + m*a(.75)

that really dosn't seem to fit a lot of the formulas make sence to me but i just feel like im missing something!
F(75)=75,F(81)=81, a(0.5)=0.5, a(.75)=0.75

Subtracting 1 from 2

F(81)-F(75)=ma(0.75)-m(0.5)
=> 81-75=0.75m-0.5m

Solve for m
 
yay dumb algebra on my part! the coefficient of friction is no problem.....surprisingly =)

thanks to u who are spending their time helpin so much, personaly i get most of it i just need a few points in the right direction!
 

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