Question about solving for Fk with work

[borderline113]

Homework Statement

A 2kg block slides from point b to point c, it has a velocity of 4 m/s at point b and a velocity of 0 at c. Distance between b and c is 3.0 m
a.)What is the coefficient of kinetic friction on the horizontal surface?

Homework Equations

Our solution sheet shows
Δk=w , N=mg

0-1/2mv²=μNd

The Attempt at a Solution

I've been working around with the problem, I got that Δk=1/2mv², but I'm not sure where to go from there. Any help would be appreciated. Test tomorrow.

 

[cepheid]

Welcome to PF,

Review (from your class notes or whatever) the work-energy theorem, which says that the work done on an object is equal to its change in kinetic energy. The change in kinetic energy is obviously equal to the difference between the final value and the initial value. You can compute both of those values using the information given.

 

[borderline113]

Welcome to PF,

Review (from your class notes or whatever) the work-energy theorem, which says that the work done on an object is equal to its change in kinetic energy. The change in kinetic energy is obviously equal to the difference between the final value and the initial value. You can compute both of those values using the information given.

Okay so it really should read 1/2mv_f²-1/2mv₀²=w however though since the final velocity is 0, the first remains 0. But then where are they getting -u_kNd from?

 

[cepheid]

Okay so it really should read 1/2mv_f²-1/2mv₀²=w however though since the final velocity is 0, the first remains 0. But then where are they getting -u_kNd from?

The change in kinetic energy is equal to the work done. What is the definition of work?

 

[borderline113]

The change in kinetic energy is equal to the work done. What is the definition of work?

Work is the energy transferred to an object. So is that the force of friction f_k=μ_kN?If so then where is the d coming from? Is it the force done over the distance it travels?

 

[cepheid]

It is true that when work is done, energy is transferred. But that is not the definition of work. Go back to your notes and look up the definition.

 

[borderline113]

Work is equal to the force over the displacement?

 

[cepheid]

Work is equal to the force over the displacement?

That's not quite right, which is strange, because this is a really simple thing to look up, even if it is not in your notes. Are you just guessing?

Not only that, but it would have covered when the concept of work was first introduced to you. (You can't just starting talking about "work" without stating clearly what it is).

 

[borderline113]

To be honest, I'm sure people complain about people having terrible physics teachers here, I don't doubt that. But this guy is honestly amazing to even be allowed to teach, not to mention that the physics department at my university is on its 10th year on probation... A 46% in the class was an A+ last quarter. It's really just sad to be honest. The first day he tried to convince us that Physics ISN'T a math course... and refuses to do any math in class because of this, add this all to the fact that the book is written by him...

 

[cepheid]

To be honest, I'm sure people complain about people having terrible physics teachers here, I don't doubt that. But this guy is honestly amazing to even be allowed to teach, not to mention that the physics department at my university is on its 10th year on probation... A 46% in the class was an A+ last quarter. It's really just sad to be honest. The first day he tried to convince us that Physics ISN'T a math course... and refuses to do any math in class because of this, add this all to the fact that the book is written by him...

Okay, so Google is your friend. You can look up "work" on Wikipedia, or even right here in the Physics Forums library, where we have articles on a variety of physics concepts. I mean, I could just tell you, but I sort of expected you to already know, and in the time you have spent lamenting the terrible quality of your physics program, you could have found out.

Note, when you said "work is equal to the force over the displacement?", I assumed that by "over", you meant, "divided by", since that is what most people mean when they say it. If that's what you meant, then it is wrong. But I just realized it is possible you meant something else.

 

[cepheid]

The first day he tried to convince us that Physics ISN'T a math course... and refuses to do any math in class because of this, add this all to the fact that the book is written by him...

It's true that Physics isn't math. But it is a science that uses mathematical models to describe nature. Therefore, mathematics is the main tool of physics, and one could even describe it as the language of physics.

If what you say is true, your teacher's stance is ridiculous.

 

[borderline113]

Okay, I looked up in the library that you mentioned(Thanks that's a lot of help, goning to use that later.) So I found that work is equal to the dot product of force and displacement. I'm assuming after that I'll just place in μ_kN for force, and then from there replace mg for N. I ended up with a final answer of .27 using 9.81 for gravity. Does that sound about right.

Also I realize I'm probably being needy, but you've helped me a ton and I can't thank you enough.

 

[cepheid]

Okay, I looked up in the library that you mentioned(Thanks that's a lot of help, goning to use that later.) So I found that work is equal to the dot product of force and displacement. I'm assuming after that I'll just place in μ_kN for force, and then from there replace mg for N. I ended up with a final answer of .27 using 9.81 for gravity. Does that sound about right.

Also I realize I'm probably being needy, but you've helped me a ton and I can't thank you enough.

Yeah that's all exactly right. Work is the dot product of force and displacement. For the simple case where these two vectors point in the same direction, this reduces to the work being equal to the magnitude of the force times the magnitude of the displacement. For this reason, people sometimes casually say "work = force*distance", it being understood that this applies in the simple 1D case.

The frictional force is μ_kN, and the normal force is just equal (in magnitude) to the weight in this situation, since all vertical forces must balance each other. So, barring any arithmetic errors, I would say that you've got it.