# Loss of mechanical energy due to friction - ?

• nukeman
Thanks!Just to be clear on some things. What does "U" represent ?U represents the potential energy of the 6kg box and 8kg box. You are correct in that the total energy is the sum of the potential and kinetic energies of the two boxes. Thanks for your help!

#### nukeman

Loss of mechanical energy due to friction - ??

## Homework Statement

Here, ill post the problem as a picture below, so you have all the info.

## The Attempt at a Solution

I am having trouble with a:

Since I don't have to worry about angles, and that the coeff of friction is not given, this is what I came up with, but I really think its wrong.

(8)(9.8) (1.5) = 117.6 J lost ?

Calculate the total energy of the system before anything happens. It will be all potential since nothing's moving. Call this your initial energy.

Afterwards calculate the total energy by adding the new potential energy and kinetic energy of the system together. Call this your final energy.

Find the difference between the two by subtracting them. This is the energy lost.

nukeman said:

## Homework Statement

Here, ill post the problem as a picture below, so you have all the info.

## The Attempt at a Solution

I am having trouble with a:

Since I don't have to worry about angles, and that the coeff of friction is not given, this is what I came up with, but I really think its wrong.

(8)(9.8) (1.5) = 117.6 J lost ?

That energy loss is only the loss of Graviational Potential Energy. But the system has gained some Kinetic Energy - presumably not the whole 117.6 J.
Any discrepancy will be the energy lost to friction.

aftershock said:
Calculate the total energy of the system before anything happens. It will be all potential since nothing's moving. Call this your initial energy.

Afterwards calculate the total energy by adding the new potential energy and kinetic energy of the system together. Call this your final energy.

Find the difference between the two by subtracting them. This is the energy lost.

im not sure how to do this. What would the formula I should be using?

nukeman said:
im not sure how to do this. What would the formula I should be using?

At first nothing's moving, all your energy is gravitational potential energy which is mgh right?

The lowest point in this problem is 1.50m below where the block originally is so I'd call that point height zero. So by doing mgh of the 6kg block you have initial energy.

When the block moves down 1.50m h=0 so no potential, but it is moving at some given speed so 0.5mv2 All you need to do is take the difference of these two quantities, m even cancels.

Now you may be thinking what about the 8kg block, doesn't it have some potential energy? It does but since the height never changes the potential energy never changes so when you take the difference it would just cancel out.

Thanks for your help...But I don't quite understand what you said here:

When the block moves down 1.50m h=0 so no potential, but it is moving at some given speed so 0.5mv2 All you need to do is take the difference of these two quantities, m even cancels.

Is there other ways to solve this? Thanks again!

nukeman said:
Thanks for your help...But I don't quite understand what you said here:

When the block moves down 1.50m h=0 so no potential, but it is moving at some given speed so 0.5mv2 All you need to do is take the difference of these two quantities, m even cancels.

Is there other ways to solve this? Thanks again!

Have you learned about potential and kinetic energy?

Let's figure out the initial energy. PE + KE since we have no movement there is no KE right? 0.5mv2 v=0 so there's none.

So what's the potential energy of the 6kg block?

When calculating the final energy there is no potential since h=0 and potential= mgh

So what's the kinetic energy when the blocks are moving 2.20m/s?

JUST started potential and kinetic today...

Is the potential energy of the 6kg box;

(6)(9.8)(1.5) = 88.2 J?

And then so I go;

.5(6)(2.20)^2 = 43.56 J

?

nukeman said:
JUST started potential and kinetic today...

Is the potential energy of the 6kg box;

(6)(9.8)(1.5) = 88.2 J?

And then so I go;

.5(6)(2.20)^2 = 43.56 J

?

Very close. The potential energy of the 6kg box is correct, however the 8kg also has some PE although you don't know what it is, just call it U

So initially the energy is 88.2 +U

The second part is also very close you've correctly calculated the kinetic energy of the 6 kg box, but what about the 8kg box?

43.56 + kinetic energy of the 8kg box + U since the 8kg box didn't change height and still has the same potential energy.

Next you need to subtract them:

43.56 + KE of 8kg box + U - (88.2 + U)

HINT: You don't need to know what U is, it just cancels

Ohhh.

Just to be clear on some things. What does "U" represent ?

So, am I just adding together the KE of 6kg box and 8 kg box, and subtracting the 43.56 ?

nukeman said:
Ohhh.

Just to be clear on some things. What does "U" represent ?

The potential energy of the 8kg box, I don't think you have enough info to figure out what it is, but it won't matter since it cancels.

So, am I just adding together the KE of 6kg box and 8 kg box, and subtracting the 43.56 ?

Not quite. You're adding together the KE of the 6 and 8kg box and subtracting from 88.2

Little lost here.

The KE of 6kg box is 88.2 J - Now you say add that to the KE of the 8kg box ?

is it:

43.56 + 117.6 - 88.2 = 72.96J lost?

If that's correct, then using that answer, how am I to figure out the coefficient of kinetic friction between the 8kg box and the tabletop?

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nukeman said:
is it:

43.56 + 117.6 - 88.2 = 72.96J lost?

If that's correct, then using that answer, how am I to figure out the coefficient of kinetic friction between the 8kg box and the tabletop?

Can't be.

You "lose" 88.2 Joules of energy due to the change in Potential energy of the 6 kg block.

You "gain" some of that due to the increase in Kinetic energy of both blocks. That kinetic energy must be less than 88.2 J or the supplied data is inconsistent.

Once you have the amount of energy lost, you can find the friction Force, because Force x distance gives the amount of energy involved.

Once you know the size of the force, you can use the weight of the 8kg block to calculate the coefficient of friction.

NOTE: your figure of 43.56 is wrong.

0.5 x 6 x 2.22 is only 14.52

I suspect you calculated (0.5 x 6 x 2.2)2
The most common way to make that mistake is a failure to understand how your calculator does things.