Help with conservation of Mechanical Energy

AI Thread Summary
The discussion focuses on the conservation of mechanical energy in various scenarios involving falling objects and jumping athletes. For the bird dropping a fish, the initial kinetic energy was miscalculated, leading to an incorrect final speed of 10.29 m/s; the correct speed upon hitting the water is 20.74 m/s. The diver's calculations for speed at different heights were confirmed as correct, with speeds of 9.9 m/s and 14 m/s respectively. The runner's jump height was estimated incorrectly, with the correct method involving energy conservation yielding a height of approximately 0.184 meters. Overall, the thread emphasizes the importance of accurately applying the principles of kinetic and potential energy in solving these physics problems.
austin1250
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Alright I have a few problems here.

1). A bird is flying with a speed of 18 m/s over water when it accidentally drops a 2 kg fish. If the altitude of the bird is 5.40 meters and friction is disregarded, what is the speed of the fish when it hits the water?

My work: I don't know if I did this one right so if you could check that would be great.
I did PE= (2kg)(9.81m/s^2)(5.4m) = 106 PE
So I made KE at the point of the water 106 KE and 0 PE (potential energy.)

Then I did 106 = 1/2 * 2kg * v^2
I got it to equal v = 10.29 m/s is that correct?

2) A 755N Diver drops from a 10 m boarding above the waters surface. Find the divers speed 5 meters above the water. Then find the diver's speed just before it hits the water.

ok here i got 9.9 m/s for the part 5 meters above water is that correct? and then I got 14 m/s when striking the water?

(Part B) If the same diver leaves the board with initial upward speed of 2.0 m/s find the divers speed when striking the water.

I worked it out until here now I don't know what to do next

3) An Olympic runner leaps over a hurdle. If the runners initial vertical speed is 2.2 m/s how much will the runners center mass be raised during the jump?

Work: I am not really sure what to do here but is the answer .5 meters up?
I don't know if this is a correct equation, but I did V^2 = mgh

4) A pendulum bob is released from some initial height such that the speed of the bob at the bottom of the swing is 1.9 m/s. what is the initial height of the bob?

Ok here i don't know what to do really. I think it goes along the lines of one above, but I am not sure if that was right either.
 
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Q1 No, Whether before dropped the fish was moving ? If yes was it having kinetic energy ?
 
will someone show me that Joule/Kilogram is equivalent to meter/second? and how can I compute the highest and lowest KE and PE there are no mass, and velociy given?
 
Welcome to PF.
A joule = kg - m²/s²
http://en.wikipedia.org/wiki/Joule

Take away kg and you have m²/s² not m/s.

As to the highest and lowest KE and PE, what is the problem you are trying to solve?
 
Question 1 is incorrect because you set the KEi as 0 when it is actually 324 J.. Okay, so you know that KEi= (1/2)mvi2.. so KEi= (1/2)(2 kg)(182) giving you 324 J.. & due to the law of conservation of energy, energy can be neither created or destroyed so the initial mechanical energy is equal to the final mechanical energy. Subsitute so you know that PEi + KEi = PEf + KEf.. Now subsitute the information you have in... 105.948 + 324 = 0 + (1/2)(2)vf2.. the final answer is 20.73518748 m/s

Question 2 is correct!

Hope that helped!
 
Last edited:
On Q5, set KE=PE
1/2mv^2=mgh
The masses on both sides cancel out
1/2v^2=gh
Then plug in your numbers
1/2(1.9m/s)^2=(9.81m/s^2)(h)
1.805=9.81h
Then divide both sides by 9.81 and it gives you h=.184m
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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