Finding coefficient of friction and the mass

[perez1028]

Homework Statement

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.

Homework Equations

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

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 don't think that is right I am not really sure where the 2 different accelerations and forces come into play...

 

[LowlyPion]

Homework Statement

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.

Homework Equations

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

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 don't think that is right I am 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.

 

[perez1028]

well not really sure what your saying by that but with uk*F = m i can get the mass but I am really not sure how to use the 75N to first get uk or the mass... that's what I am missing

 

[rock.freak667]

So we know that F_netma_net.

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

F_net=F'-F
so that ma_net=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.

 

[perez1028]

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 sense to me but i just feel like I am missing something!

 

[rock.freak667]

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 sense to me but i just feel like I am 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

 

[perez1028]

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!