Energy, work, 2 masses on incline

[FurFur]

Homework Statement

Two blocks with masses m1=4.0kg and m2=5.0kg are connected by a light rope and slide down a frictionless wedge. The angle of the side on which m2 is moving is 53 degrees, and the angle of the side on which m1 is moving is 37 degrees. Basically the system looks like a triangle with the masses sliding down the two sides. If the system starts from rest, what is the speed of m2 after it has moved 40.0cm along the incline?
GIVEN: m1=4.0kg m2=5.0kg d2(displacement of mass 2)=40.0cm

Homework Equations

Ek= 1/2mv^2 Eg=mgh

The Attempt at a Solution

I know that in this situation, the acceleration is going to be towards m2 so m2 is going to lose gravitational potential energy and gain kinetic. I know that m1 is going to gain gravitational potential energy and lose kinetic energy. I also know that the energy is conserved. I could calculate Eg for m2 but I am not sure if i would use 40.0cm for the height.

 

[alphysicist]

Homework Statement

Two blocks with masses m1=4.0kg and m2=5.0kg are connected by a light rope and slide down a frictionless wedge. The angle of the side on which m2 is moving is 53 degrees, and the angle of the side on which m1 is moving is 37 degrees. Basically the system looks like a triangle with the masses sliding down the two sides. If the system starts from rest, what is the speed of m2 after it has moved 40.0cm along the incline?
GIVEN: m1=4.0kg m2=5.0kg d2(displacement of mass 2)=40.0cm

Homework Equations

Ek= 1/2mv^2 Eg=mgh

The Attempt at a Solution

I know that in this situation, the acceleration is going to be towards m2 so m2 is going to lose gravitational potential energy and gain kinetic. I know that m1 is going to gain gravitational potential energy and lose kinetic energy.

That's not correct for $m_1$; it will gain potential energy, but if it starts from rest, how can it lose kinetic energy?

I also know that the energy is conserved. I could calculate Eg for m2 but I am not sure if i would use 40.0cm for the height.

The $h$ in the gravitational potential energy formula is the position in the vertical direction; so how much it has moved in the horizontal direction does not appear in $h$ at all. Do you see how to set it up now? What do you get?

 

[FurFur]

I figured the kinetic energy for m1 is going to decrease and be converted to potential energy.
For Eg of m2 i could use the formula Eg=mgh. The block slides down 40.0cm on an angle of 53degrees so the horizontal height would be 0.4m sin53= 0.32m. I substituted that into Eg=mgh
= 5 x 9.8 x 0.32
= 15.68J

 

[alphysicist]

I figured the kinetic energy for m1 is going to decrease and be converted to potential energy.

Since the smallest value for the kinetic energy is zero, there is no way for it to decrease. Both objects are speeding up, and so both of their kinetic energies are increasing.

For Eg of m2 i could use the formula Eg=mgh. The block slides down 40.0cm on an angle of 53degrees so the horizontal height would be 0.4m sin53= 0.32m. I substituted that into Eg=mgh
= 5 x 9.8 x 0.32
= 15.68J

That's looks like the right magnitude, but you'll need to be careful with the sign. Most people would say that the original height of m2 was h=0, and so this height is h=-0.32 m (since it's sliding downwards).

But I've seen other people assemble their equation a bit differently, so what do you get for the entire energy equation?

 

[FurFur]

h=0 is my reference point.
for the kinetic energy of m2 i got 15.68 J
Eg=mgh
= 5 x 9.8 x 0.32
= 15.68J
but i have no clue where to go from there:( i need to somehow involve m1 into this and the law of conservation of energy.

 

[alphysicist]

h=0 is my reference point.
for the kinetic energy of m2 i got 15.68 J
Eg=mgh
= 5 x 9.8 x 0.32
= 15.68J
but i have no clue where to go from there:( i need to somehow involve m1 into this and the law of conservation of energy.

No, that's not right for the kinetic energy of m2. You have to deal with both masses together because they affect each other.



Since you said energy is conserved, that means that the total energy of both masses at the starting point will be equal to the total energy of both masses at the final point, $E_i = E_f$.

So at the beginning, when both masses are at rest, what is the total energy of the system? To figure this out, you've already listed the types of energies possible, so what is the kinetic energy and potential energy of m1? what is the kinetic energy and potential eneryg of m2? Remember that you already have the masses, so all you need is the speed and height of each one.


Then do the same for the final point: what is the kinetic and potential energy of each mass at the final point? The speed at the final point is the unknown, so all you need is the height of each mass at each of their final points.


Altogether you'll have eight energy terms, but four of them should be zero.

 

[FurFur]

okay but arent the speed and the height of both masses zero since they start at rest (v=0) and my reference point is 0 (h=0)? substituting zero to the equation will get me zero...that doesn't seem right.

 

[alphysicist]

okay but arent the speed and the height of both masses zero since they start at rest (v=0) and my reference point is 0 (h=0)? substituting zero to the equation will get me zero...that doesn't seem right.

No, that's perfect! Remember you're not setting a whole equation to zero (yet), its just the initial energy. So the initial energy is zero, or:

$E_i$ = (initial KE of $m_1$ ) + (initial PE of $m_1$) + (initial KE of $m_2$) + (initial PE of $m_2$) = 0

So now calculate the four energy terms for $E_f$ (just like you did for $E_i$; and then to get your equation, you'll set

$$E_i = E_ f$$

 

[FurFur]

Oh...okay. Yea I tried it and surely enough I had 8 terms and 4 were zero.
I got the height using 0.4sin53= 0.32 and the and speed of mass 1 using: Eki+Egi=Ekf+Egf
vi=2.5m/s
i substituted that into the 8 term equation below:
1/2m1v1+1/2m2v2 +m1gh +m2gh= 1/2m1v1' +1/2m2v2' +m1gh' +m2gh'
0 +0 +0 +15.68= 12.5 +2.5v2^2' + (-12.544)
v2'=2.5m/s
Does make sense? would it be correct? I hope i didn't forget to to accoutn for anything else.

 

[FurFur]

account*

 

[alphysicist]

Oh...okay. Yea I tried it and surely enough I had 8 terms and 4 were zero.
I got the height using 0.4sin53= 0.32 and the and speed of mass 1 using: Eki+Egi=Ekf+Egf
vi=2.5m/s
i substituted that into the 8 term equation below:
1/2m1v1+1/2m2v2 +m1gh +m2gh= 1/2m1v1' +1/2m2v2' +m1gh' +m2gh'
0 +0 +0 +15.68= 12.5 +2.5v2^2' + (-12.544)
v2'=2.5m/s
Does make sense? would it be correct? I hope i didn't forget to to accoutn for anything else.

It's almost there, but there are a few problems.

Your energy equation and the equation with values is:

1/2m1v1+1/2m2v2 +m1gh +m2gh= 1/2m1v1' +1/2m2v2' +m1gh' +m2gh'

0 +0 +0 +15.68= 12.5 +2.5v2^2' + (-12.544)

The left hand side is okay; in particular, you have set the original height of m1 to be zero, and the final height of m2 to be zero. That's fine, we just have to keep track of it.

On the right side, there are a couple of problems. You don't have the kinetic energy term for m1. Also, you have two potential energy terms for m1, one with a positive sign and one with a negative sign. To decide which is which, since m1 started at h=0, does it end up higher or lower than it began? If it's higher, the final h will be positive (and so will the final potential energy of m1); if it's lower, h will be negative.

(Also, I don't think the number 12.5 is correct; I think you might have used the wrong angle for m1.)

So you still need to find the following terms:

$$\begin{align} \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 +m_1gh_1 +m_2gh_2= \frac{1}{2}m_1v_1'^2 +\frac{1}{2}m_2v_2'^2 +m_1gh_1' +m_2gh_2'\nonumber\\ 0+0+0+15.68=(\mbox{kinetic energy term for m1?})+2.5v_2'^2+(\mbox{potential energy term for m1?}) + 0\nonumber \end{align}$$

Once you get those last two terms in with the right variables and values, you'll be able to find the speed.

 

[FurFur]

im getting 2.5m/s for my kinetic energy for m1final. For the Eg for m1final I'm getting -12.544. I am not sure if these numbers are right though since in mgh i made g= -9.8.

 

[FurFur]

oops mistake...for my kinetic energy for m1final I'm getting 12.5 and for the Eg of m1 final I am getting zero

 

[alphysicist]

im getting 2.5m/s for my kinetic energy for m1final. For the Eg for m1final I'm getting -12.544. I am not sure if these numbers are right though since in mgh i made g= -9.8.

The g always stands for the magnitude 9.8 m/s^2; you don't make it negative. So then the final PE for m1 is positive (since the final h for m1 is positive).

But how did you get the number 12.544? Did you use the wrong angle? If you're still getting 12.54 J, please post the actual numbers that you used. (And remember, m1 is on a 37 degree incline.)

For the kinetic energy of m1, you can't find the actual number yet; you don't know how fast m1 and m2 are going. That's what you are trying to find in this problem. So the KE term for m1 should have the unknown variable v in it.

 

[FurFur]

okay I think I am doing something incredibly wrong in my angles. When i try to get the height of the triangle I am doing 0.40m x sin 53 which gives me 0.32. The height occurs several times in the 8 term equation which is why my answer keeps coming wrong. I am not sure what to do about the height.

 

[FurFur]

I wish i could upload an image. This is what the triangle looks like. 0.40 is the hypotenuse of the triangle with the angle 53degrees when you split this triangle into two. That is why I am getting 0.32m as my height. Something's wrong...

 

[alphysicist]

okay I think I am doing something incredibly wrong in my angles. When i try to get the height of the triangle I am doing 0.40m x sin 53 which gives me 0.32. The height occurs several times in the 8 term equation which is why my answer keeps coming wrong. I am not sure what to do about the height.

The mass m2 is on the 53 degree angle incline; so it moves downards a vertical distance of 0.32 meters.

The mass m1 is on the 37 degree angle incline; so how far vertically does it move upwards? You would do it the same way, just replacing the angle.

 

[alphysicist]

This is how I was visualizing it, with m1 on the left and m2 on the right. Is this not correct?

http://img515.imageshack.us/img515/2570/inclinend2.jpg

 

[FurFur]

well the angles are switched. 53degrees is on the left with m2 and 37degrees is on the right with m1

 

[FurFur]

to get the height I am doing 0.40sin53. So when i split the triangle that you've drawn, into 2 the height of both is the same therefore its 0.32.

 

[alphysicist]

well the angles are switched. 53degrees is on the left with m2 and 37degrees is on the right with m1

Okay, good; left and right won't matter.

So m2 goes down the incline a distance of 0.4m, so it moves vertically

m2--> 0.4 sin(53 degrees)=0.32 meters

as you've already found.

For m1, it moves up the incline a distance of 0.4 m, so it moves vertically

m1--> 0.4 sin(37 degrees)

and that is the height that goes in the PE term for m1.

to get the height I am doing 0.40sin53. So when i split the triangle that you've drawn, into 2 the height of both is the same therefore its 0.32.

You don't want to split the triangle into two; the block are not moving down the entire length of the incline; they're just going partway. Does that make sense?

 

[alphysicist]

Look at this picture, which shows the initial and final position of each mass.

http://img378.imageshack.us/img378/4290/incline2lw7.jpg

From that picture, the red lines are the heights that we are trying to find. You've already found the h2 for m2, but the h1 for m1 will be different.

 

[FurFur]

Oh..okay that makes more sense. So the height of mass 1 is 0.24m. Now I need to get the final speed of mass 1. this is what i did:
Ek1 +Eg1= Ek2 +Eg2
1/2mv1^2 +mgh1= 1/2mv2^2 +mgh2
1/2(4)(0) + (4)(9.8)(0)= 1/2(4)v2^2 +(4)(9.8)(0.24)
0 +0 = 1/2(4)v2^2 + 9.41
-9.41/0.5(4)= v2^2
-4.71=v2^2
i can't take the square root of a negetive number and that is what's keeping me from getting the final speed of mass 1.

 

[alphysicist]

Oh..okay that makes more sense. So the height of mass 2 is 0.32m. Now I need to get the final speed of mass 2. this is what i did:
Ek1 +Eg1= Ek2 +Eg2
1/2mv1^2 +mgh1= 1/2mv2^2 +mgh2
1/2(4)(0) + (4)(9.8)(0)= 1/2(4)v2^2 +(4)(9.8)(0.32)
0 +0 = 1/2(4)v2^2 + 12.544
-12.544/0.5(4)= v2^2
-6.272=v2^2
i can't take the square root of a negetive number.

That equation won't work; the only equation we can work with is the full energy equation (the eight-term one if we keep all those zero terms for now).

You have three questions to answer before you're finished writing it down:

What is h1, the height of mass 1 (from the picture I uploaded)?

Once you know h1, what is the numerical value for the PE term for m1? (It's formula is $m_1 g h_1$; what number is that?)

The final question, what is the formula (we can't find the number yet) for the final kinetic energy of m1?

Once you put in the answers to the second and third questions on the right hand side of the full energy equation (back in post #11), you'll be able to start solving for v. What do you get?

 

[FurFur]

I edited my post...im not sure what to do about the negetive number

 

[alphysicist]

I edited my post...im not sure what to do about the negetive number

The speed of m1 and m2 are both v, the unknown quantity that we are trying to find.

Just like you put in (1/2) 5 v^2 in for the final kinetic energy of m2, you'll put in (1/2) 4 v^2 for the final kinetic energy of m1. Since they are connected by a rope, they have to have the same final speed v.

 

[FurFur]

o okay okay...so to get the final speed of m1 i don't do this:
Ek1 +Eg1= Ek2 +Eg2
1/2mv1^2 +mgh1= 1/2mv2^2 +mgh2
1/2(4)(0) + (4)(9.8)(0)= 1/2(4)v2^2 +(4)(9.8)(0.24)
0 +0 = 1/2(4)v2^2 + 9.41
-9.41/0.5(4)= v2^2
-4.71=v2^2

instead for the final speed of m1 i jus leave it at (1/2) 4 v^2 and substitute that into the 8 term formula?

 

[FurFur]

As well, the 2 heights for m1 are : 0m and 0.24m
to heights for m2 are : 0m and 0.32m.
Im not sure, for both masses, which height is initial and which is fiinal. Both can't start at zero since m1 moves up and m2 moves down.

 

[alphysicist]

o okay okay...so to get the final speed of m1 i don't do this:
Ek1 +Eg1= Ek2 +Eg2
1/2mv1^2 +mgh1= 1/2mv2^2 +mgh2
1/2(4)(0) + (4)(9.8)(0)= 1/2(4)v2^2 +(4)(9.8)(0.24)
0 +0 = 1/2(4)v2^2 + 9.41
-9.41/0.5(4)= v2^2
-4.71=v2^2

instead for the final speed of m1 i jus leave it at (1/2) 4 v^2 and substitute that into the 8 term formula?

That's right.

As well, the 2 heights for m1 are : 0m and 0.24m
to heights for m2 are : 0m and 0.32m.
Im not sure, for both masses, which height is initial and which is fiinal. Both can't start at zero since m1 moves up and m2 moves down.

Well, you could start them both at zero--then the final height of m2 would just be a negative number.


(However, in the last version of the full formula (post #11), you had already chosen the initial height of m1 to be zero, and the final height of m2 to be zero, which is fine.)

So going back to the version in post #11, which was:

$$\begin{align}\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 +m_1gh_1 +m_2gh_2= \frac{1}{2}m_1v_1'^2 +\frac{1}{2}m_2v_2'^2 +m_1gh_1' +m_2gh_2'\nonumber\\0+0+0+15.68=(\mbox{kinetic energy term for m1?})+2.5v_2'^2+(\mbox{potential energy term for m1?}) + 0\nonumber\end{align}$$

After plugging in your expression (on the right side) for the m1 kinetic energy, and the numerical value of the m1 potential energy, what do you get?

 

[FurFur]

After I've substituted the numbers I get:
15.68 - 9.408= 2v1^2' + 2.5v2^2'
im not sure what to do with the right side mathematically

 

[alphysicist]

After I've substituted the numbers I get:
15.68 - 9.408= 2v1^2' + 2.5v2^2'
im not sure what to do with the right side mathematically

How are v1 and v2 related? Can you rewrite it so that you only get one unknown?

 

[FurFur]

well v1 and v2 must be the same.
15.68 - 9.408= 2 (v1^2' + 1.25v2^2')
this still leaves me with 2 unknowns..but both unknowns are equal..hmm

 

[alphysicist]

well v1 and v2 must be the same.
15.68 - 9.408= 2 (v1^2' + 1.25v2^2')
this still leaves me with 2 unknowns..but both unknowns are equal..hmm

If they are both the same (and they are), give them the same name. Call them both v. Then you have only one unknown.


Or, if you want to look at it another way, you know that v1=v2, so replace v1 with v2 in your equation.

 

[FurFur]

oh right! Okay so the equation is:
15.68 - 9.408= 2v^2 + 2.5v^2
6.272= 4.5v^2
6.272/4.5 = v^2
1.39]square root = v
1.18 m/s= v
that seems right.

 

[alphysicist]

oh right! Okay so the equation is:
15.68 - 9.408= 2v^2 + 2.5v^2
6.272= 4.5v^2
6.272/4.5 = v^2
1.39]square root = v
1.18 m/s= v
that seems right.

That looks right to me.

So the interpretation of the energy formula might be, m2 loses potential energy, which is transformed into increasing potential energy for m1 and also increasing kinetic energy for m1 and m2.