x Turn on thread page Beta
 You are Here: Home >< Maths

# Some more M3 help. Slopes and Energy watch

1. Ok, so if you have a rough slop with a particle at one point, and the falling to another. You would say:

Energy at Point A: KE: 0.5mv^2 + GPE: mgh

Energy at point B : KE: 0.5mv^2 + GPE: mgh + WD against Friction

And then you would put these equal to each other and solve.

But in this question:

I would equate it as:

Energy at A = 0

Energy at B :

So I thought it would be EPE + GPE + WD (against Friction) would be = 0

But, I see in the answer that lambda is supposed to be 66. And when I put 66 in, I get 0 WITHOUT plusing the WD. So what have I done wrong?

What I think is that the WD is included in the EPE, but ill let someone explain.

Thanks
2. (Original post by 2710)
What I think is that the WD is included in the EPE, but ill let someone explain.
The work done against friction is simply nothing more than the work required to move the particle the distance moved against the restraining force of friction. i.e. it's simply W = Fd, where F = frictional force = , and d = distance moved.
3. (Original post by Kolya)
The work done against friction is simply nothing more than the work required to move the particle the distance moved against the restraining force of friction. i.e. it's simply W = Fd, where F = frictional force = , and d = distance moved.
I know that WD is FxD, but it is also conserved in the energy conservation.

So how come when a particle is released is on a rough slope, you do:

1/2 (m)(V)^2 + WD + -(mgh) = 0 (assuming datum line from the top point)

Ie, you include the Work Done in the formula. Whereas in this question, they are saying that:

-(mgh) + EPE = 0..... so wheres the WD?

Shouldnt it be:

- (mgh) + EPE + WD = 0 Like we did with the first one? Its exactly the same, but just with a band tying the particle.

Thanks
4. (Original post by 2710)

So I thought it would be EPE + GPE + WD (against Friction) would be = 0

But, I see in the answer that lambda is supposed to be 66. And when I put 66 in, I get 0 WITHOUT plusing the WD. So what have I done wrong?

What I think is that the WD is included in the EPE, but ill let someone explain.

Thanks
Another way of thinking about it is , initally it has GPE and then moving down the slope its losing energy to friction and when youre at B the amount of energy left s only due to EPE
so
5. (Original post by rbnphlp)
Another way of thinking about it is , initally it has GPE and then moving down the slope its losing energy to friction and when youre at B the amount of energy left s only due to EPE
so
Um...thats exactly wut I done, except that you took the datum line from point B.

So according to you

EPE + WD = GPE

Now if you put the values in:

which is wrong...
6. (Original post by 2710)

[
EPE my friend is not
its
7. (Original post by rbnphlp)
EPE my friend is not
its
**** lol, omg, let me go check if this is right then. Ill be back later >____>

EDIT: I can't believe all of this was because of that stupid mistake >___> Thanks!

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 24, 2010
Today on TSR

### Any tips for freshers...

who are introverts?

### More snow?!

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE