Hello guys. I am new to this forum,so nice to meet you all. Instead of chit-chat,I will get to the point at once. I am having problems regarding a Physics problem. Care to solve it for me?
The question is as follows :-
A stone is dropped from a height h.At a particular instant,the potential energy was 9 times the kinetic energy. When the velocity gets doubled,what is the new ratio of potential energy to kinetic energy?
Please reply as fast as possible. When you are answering,please provide explanations wherever possible. Posting pictures to demonstrate the logic is appreciated.
Thank you for your cooperation.
206Use Law of Conservation of Energy for solving such problems..
Just remember that
Initial total energy = Final total energy
or, Initial PE + Initial KE = Final PE + Final KE
Arkadyuti Banerjee Can you please clarify? Thanks for the reply anyways
337I have tried to solve this problem but I am not sure how much correct I am. I request all the forum members to report any flaw they find in my procedure:
Initial PE= mgh
Initial KE = 0
Let velocity at that point be v.
So KE at that point= 12mv2
PE at that point = 9 x 12mv2
As Sayantan sir said, KE + PE at any point is the same.
So KE+ PE at that point = Initial PE + Initial KE.
=> 12mv2+9 x 12mv2=mgh
=> 5v2=gh
=> 4v2=2gh5
Now let the height of the stone when velocity gets doubled be H.
So, 12m(2v2)+mgH=mgh
=> 12m4v2+mgH=mgh
Replace 4v2 with 2gh5
Then H comes out to be = h2
Therefore, 4 x 12 mv2+12mgh=mgh
Therefore, the new ratio is:40.5=8:1
Thanx...
Sayan Sinha Sorry... I gave the ratio between KE and PE. The question asks for PE:KE. So the answer, according to me, should be 1:8.
Thanx...Sayantan Hazra The procedure of solving is fine; could have been done in a shorter way. There are two mistakes here:
Mistake 1: 5v^2 = gh; chk d nxt step
Mistake 2: You are asked to find the ratio of PE to KE at that instant; not KE to the total PE, which is what you have found out.Sayan Sinha Yes sir. It should have been 4gh/5. I forgot to take the square of 2.
Sir, I have calculated the height of the ball at the instant when velocity gets doubled, and it comes out to be h/2 in my solution. But, I had made a mistake in calculating 4v^2. Therefore, the new H= 3h/5. Upon further calculation (similar steps as the solution I had given), the answer comes out to be 3:20...
Thanx...Arkadyuti Banerjee Thanks guys for helping me. Looks like I had done it right only. I just needed to think straight. Thanks for your response.Arkadyuti Banerjee Thank you Sir for helping me! How to close this thread BTW?206i still don't think you have arrived at the ans.....
3:20 is incorrect
Sayan Sinha Sir, may I know why?Sayantan Hazra because you are making the same mistake of finding out the ratio of total PE to KE and not instantaneous PE to KE.....try to find out the PE & KE at that instant, and then find the ratioSayan Sinha Sir, but I calculated height at that instantaneous point when velocity was double. I am not comparing it to the total PE. I am equating it to total PE so that it is easy to understand the ratio derivation. Instantaneous PE+KE can be equated to mgh or 1/2 mv^2 [here v is the final velocity before touching the ground and h is the original height].
Thanx...
106The correct answer is 3:2. (I haven't really checked what has been done in ans #1 so I can't comment on any mistakes)
Solution:
PE1KE1 = 91 .... (1)
When velocity is doubled, KE is 4 times original.
=> KE2 = 4KE1 .... (2)
PE2 + KE2 = PE1+KE1 .... (3)
From (1),(2) and (3), ratio is easily calculated.