'solved' a question but don't know which one is right.

[Fishingaxe]

Homework Statement

An object with mass 2.0 kg is dropped from level A in the figure below.
a) In which of the selected levels A - L is its kinetic energy 14 J?
http://img40.imageshack.us/img40/9245/0aup.png

Homework Equations

T = 1/2m*v^2 ?
Or P = m*v ?

The Attempt at a Solution

Using the first equation I got: = 0.5 * 2 * (9.81)^2*0.1 = 9.62 J. ( I put the 0.1 at the end to find out the kinetic energy per square(0.1m)

14/9.62 = 1.455 (To find out how many squares it would take to achieve 14J)

Using the second equation: 2.0 * 9.81*0.1 = 1.962 J (Again 0.1 to find out the kinetic energy per square)

14/1.962 = 7,1355 (and how many squares it'd take to achieve 14J)

I got a right answer at this problem but he didn't tell me which one was correct and I have a test tomorrow and really need to understand which one is right. Any help is greatly appreciated.

 

[cepheid]

Hello Fishingaxe,

I can't quite read the units that tell you the grid spacing. Does it say 1.0 dm (as in decimetres)?

In any case, your approach is not correct. You need to use energy conservation to figure out the change in energy as a function of vertical distance below level A. Then you will be able to compute the kinetic energy at each level. (You don't have to actually compute it separately for each level, you can just use algebra to solve for the height where it's equal to 14 J).

 

[Fishingaxe]

Hello Fishingaxe,

I can't quite read the units that tell you the grid spacing. Does it say 1.0 dm (as in decimetres)?

In any case, your approach is not correct. You need to use energy conservation to figure out the change in energy as a function of vertical distance below level A. Then you will be able to compute the kinetic energy at each level. (You don't have to actually compute it separately for each level, you can just use algebra to solve for the height where it's equal to 14 J).

Sorry for the picture but yes it says dm. So neither of my answers were correct?

How do I figure out the change in energy like you said using "energy conservation"? is it V=mgh?

So V(potential energy)=2*9.81*1.1 ≈ 21.6 J

21.6/11=1.96/square 14/1.96 = 7.14 dm(squares) down. Is this correct?

 

[Fishingaxe]

That is the exact same answer as using the 2nd equation I had with less decimals, so if this is correct I guess it doesn't matter which one I use?

 

[cepheid]

That might be correct, although you can do this more easily without it being a two-step process (you don't have to compute the change in KE for the whole drop, and then use that to determine the change in KE per square, and then figure out the number of squares required for the energy to change by 14 J). You can do it all in one step:

Initially, we can say that the kinetic energy is 0 (the object is at rest before it starts falling). We know from energy conservation that -ΔV = ΔT (to use your notation for PE and KE respectively).

So we want ΔV = -14 J, since when the PE has been reduced by 14 J, the KE will have increased by 14 J.

mgΔy = -14 J, where Δy is the change in vertical position (y).

Solve for Δy. I get an answer of -7.14 dm. So the object has to have dropped by 7.14 squares in order to have gained 14 J of KE.

 

[Fishingaxe]

That might be correct, although you can do this more easily without it being a two-step process (you don't have to compute the change in KE for the whole drop, and then use that to determine the change in KE per square, and then figure out the number of squares required for the energy to change by 14 J). You can do it all in one step:

Initially, we can say that the kinetic energy is 0 (the object is at rest before it starts falling). We know from energy conservation that -ΔV = ΔT (to use your notation for PE and KE respectively).

So we want ΔV = -14 J, since when the PE has been reduced by 14 J, the KE will have increased by 14 J.

mgΔy = -14 J, where Δy is the change in vertical position (y).

Solve for Δy. I get an answer of -7.14 dm. So the object has to have dropped by 7.14 squares in order to have gained 14 J of KE.

That is awesome, thank you sir. Will def use that one! Always find it interesting when there are so many correct ways to calculate something :) Beautiful in a way.