Solving a Physics Problem: Calculating Distance with Kinetic Friction and Work

[rghost90]

Here's the problem:
Marie slides a carte of nails across a hardwood floor. The coefficient of kinetic friction between the carte and the floor is 0.120. The crate has a mass of 56.8kg and marie pushes with a horizontal force of 124N. If the 74.4J of total work are done on the crate, how far along the floor does it move?

I'm a lil confused. I know that the equation of kinetic friction is fk=uk * N
But i have problem in determining where i put the variables in the equation.
Im so lost can someone pleaze help me?

 

[tiny-tim]

welcome to pf

hi rghost90! welcome to pf!

(have a mu: µ )

hint: what is the definition of work done on the crate?

 

[rghost90]

Hi and thank you. The problem doesn't state the definition of the crate. Or are you asking what is the definition of work on the crate? In that case, work is the object that is in motion due to a force and a displacement right?

 

[tiny-tim]

no, i mean what is the standard definition of "work done"?

 

[rghost90]

ummmm when an object has a force or a cause, that causes its displacement?

 

[tiny-tim]

work done = force times displacement (strictly, force "dot" the displacement of the point of application of the force)

does that help? ​

 

[rghost90]

oh you meant the equation of work? yeah i get that. I am having a problem on how to find the total displacement of the crate.

 

[tiny-tim]

oh you meant the equation of work? yeah i get that. I am having a problem on how to find the total displacement of the crate.

that's the question!

Use the definition of work done to write an equation, and put the given figures into the equation …

what do you get?​

 

[rghost90]

oh ok I think I get it. So basically the equation of work is F= W*d. So I have to rearrange the equation so that its d= F*W? and that will get me my total displacement right?

 

[tiny-tim]

oh ok I think I get it. So basically the equation of work is F= W*d. So I have to rearrange the equation so that its d= F*W? and that will get me my total displacement right?

Yup!

(except it's W = Fd, and anyway you've re-arranged it wrong )

 

[rghost90]

oh lol oops. Yeah that's what I meant to say. I am such a noob at physics

 

[rghost90]

ok i got 9225.6. That isn't the final answer is it? don't I have to divide by the kinetic friction or something?

 

[tiny-tim]

You had to multiply by µ to find the force.

Once you've done that, you can forget about µ.

 

[rghost90]

oh ok. So I multiplied the force (124N) by µ (0.120). Then I plugged the total force (14.88) into the "work done" equation ( d= F*W) and then multiplied by the total work (74.7J) and got 1107.072.

 

[tiny-tim]

hi rghost90!

(just got up :zzz: …)

oh ok. So I multiplied the force (124N) by µ (0.120). Then I plugged the total force (14.88) into the "work done" equation ( d= F*W) and then multiplied by the total work (74.7J) and got 1107.072.

erm … you've done it again!

it's W = Fd.

Try again. ​