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Conservation Energy Q's
Started by sofpink
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It took me 7 minutes to solve this (I feel kind of dumb). We are looking at the problem in a scalar sense (since we are looking at speed) so this means you need to factor in both vertical velocity and horizontal velocity. The equation to use is one of the big five: v^2 (final velocity) = v^2 (initial velocity) + 2ad.
So plug in: 3m/s (dogs horizontal speed) into v^2 (initial velocity); 10m/s^2 (gravity) into a; and 4m (distance of lower cliff from upper cliff) into d.
This will turn up:
v^2 (final) = (3)^2 +2(10)(4)
v^2 (final) = 9+80
v^2 (final) = 89
v (final) = 9.43
This is the answer.
So plug in: 3m/s (dogs horizontal speed) into v^2 (initial velocity); 10m/s^2 (gravity) into a; and 4m (distance of lower cliff from upper cliff) into d.
This will turn up:
v^2 (final) = (3)^2 +2(10)(4)
v^2 (final) = 9+80
v^2 (final) = 89
v (final) = 9.43
This is the answer.
Last edited:
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Well I feel silly because I knew to use that equation but read the question as the ledge was at 4 m meaning there was a 6 m drop. Couldn't figure out what I was doing wrong. This means I need to go to bed.
Sorry I made a mistake and said vertical speed instead of horizontal speed for the dog (just corrected it).
I guess the reason why I did not use total energy = KE + PE is because this only answer the vertical component of the question. I know both are scalar, but I don't think you can factor the horizontal component into that equation (there is no potential energy in the horizontal velocity of the dog only kinetic). This is why I used that equation I mentioned in my previous post.
I guess a second opinion on the question would be helpful.
I guess the reason why I did not use total energy = KE + PE is because this only answer the vertical component of the question. I know both are scalar, but I don't think you can factor the horizontal component into that equation (there is no potential energy in the horizontal velocity of the dog only kinetic). This is why I used that equation I mentioned in my previous post.
I guess a second opinion on the question would be helpful.
LOL... I made the same mistake and then I re-read the question twice before realizing this. Looks like I need to brush up on translational motion questions.
Oh s***, I just figured out the real answer to your question on energy. You are right the energy equation applies here. However, it needs to be set up like this:
KE (initial) + PE (initial) = KE (final) + PE (final)
PE (final) = 0 since all the potential energy at the end of flight is converted to kinetic energy
KE (final) = KE (initial) + PE (initial)
[I know the set up is a little weird but it makes sense since the KE (initial) is for the horizontal component and PE (initial) is for the vertical]
v^2(final) = 2 [ 1/2 mv^2(initial) + mgh]/ m
The "m's" cancel
v^2 (final) = 2 [1/2 v^2 (initial) + gh]
now plug in the values
v^2 (final) = 2[ 1/2(3)^2 + (10)(4)]
v^2 (final) = 2[1/2( 9) + 40]
v^2 (final) = 9 + 80
v^2 (final) = 89
v (final) = 9.43
This is the best way to think of the question. Sorry about the previous post.
