ilovephysics.com :: free online physics help, tutorials, forums, classes, misconceptions in / Help with this simple WORK problem

virtuoso_735 · 2006-03-31 03:18:59

Hi everyone. I just found this forum and it looks like a good place to get some help on my physics homework. Anyways, this year, I have bad luck. I have a REALLY bad physics teacher (I'm not bashing on teachers here, but it's true). He doesn't do anything; we haven't learned ANYTHING from his class, he is lazy and disorganized, and rambles on and on about things not related to physics, and can't even solve simple problems. Thanks for letting me rant, I'm really frustrated with this teacher. He gives us some work and expects us to do it without any help.

Here's a problem I can't figure out:

You slide a crate up a ramp at an angle of 30.0° by exerting a 230 N force parallel to the ramp. The crate moves at constant speed. The coefficient of friction is 0.28. How much work have you done on the crate when it is raised a vertical distance of 1.10 m?

I'm sure it's pretty simple to you guys, but I don't know how to do it. Can someone provide me with an answer or even an explanation along with the answer? I would say I'm a pretty smart guy, so I can figure it out if I get an answer, but an brief explanation would be nice as well.

Thanks!

virtuoso_735 · 2006-03-31 03:47:10

Okay so I tried solving it but it didn't work again.

So here is what I did:

w=fd
I solved for distance using sin(30)=1.10/d
Then I plugged in d in the formula and I plugged in the f which is 230 N according to the problem and I got 104.55 but the answer doesn't match the book. Gah!

How should the coefficient of friction somehow fit in this?

Last edited by virtuoso_735 (2006-03-31 03:47:40)

Chris · 2006-03-31 17:17:07

Not a bad start. Though here are some things you need to know about work:

Work is done when a force causes a displacement. The magnitude of the work is:

$W = \vec{F_{net}} x \vec{d}$

The "x" is a cross product, which means work is only done is the direction of the displacement.

Whenever you solve any problem, always draw a picture. Here is what we have for your problem:

Now when you are asked to find out how much work is done on an object, first look in the direction of the force doing the work (in this case up the ramp.) So your displacement is up the ramp, and like you said:

$d = sin \theta$

Now you want to find the net force in that direction. In our case:

$F_{net} = 230 N - \mu N - mg sin \theta$

And since $N = mg cos \theta$

$F_{net} = 230 N - \mu mg cos \theta - mg sin \theta$

Now here is were you have to understand some physics. The problem says that the box is moving with constant velocity. That means that in the direction up the ramp, the is NO acceleration, so therefore there can be NO net force.

So,

$F_{net} = 230 N - \mu mg cos \theta - mg sin \theta = 0$

So what this means is I have done ZERO work in the direction up the ramp! But have I done work? Yes. I have done work in the direction straight up, through the distance d.

This problem is as simple as "How much work is done when I pick up a book and set it on the table 1.1 meters off the ground?"

In this problem, they just don't give you the weight of the book, so you have to do some pretty slick figurin' to find the weight.

I'll leave that to you, but I'll give you a hint: You can use the last equation up above to figure out the weight (mg).

BTW, I get 341 J as the answer, what does the book say?

Martin · 2006-04-01 01:38:37

Chris wrote:
...The magnitude of the work is:

$W = \vec{F_{net}} x \vec{d}$

The "x" is a cross product, which means work is only done is the direction of the displacement.

I think Chris misspoke. The amount of the work is given by:

$W = \vec{F_{net}}$ • $\vec{d}$

where the “•” is a dot product.

Chris · 2006-04-01 02:33:50

I most certainly did Martin. Otherwise work would be a vector, which doesn't make much sense.

Thank you for the correction.

Martin · 2006-04-01 03:06:07

Additional point to emphasize: The dot product gives the amount of work, not the magnitude of work, as the dot product can be either negative or positive.

Chris · 2006-04-07 22:51:25

I assume virtuoso_735 has either figured this one out on his own, or left.

Announcement

#1 2006-03-31 03:18:59

Help with this simple WORK problem

#2 2006-03-31 03:47:10

Re: Help with this simple WORK problem

#3 2006-03-31 17:17:07

Re: Help with this simple WORK problem

#4 2006-04-01 01:38:37

Re: Help with this simple WORK problem

Chris wrote:

#5 2006-04-01 02:33:50

Re: Help with this simple WORK problem

#6 2006-04-01 03:06:07

Re: Help with this simple WORK problem

#7 2006-04-07 22:51:25

Re: Help with this simple WORK problem

Board footer