Child going down a water slide (potential energy and energy conservation)

[Thea Woods]

Homework Statement

A 56 kg child slides down a water slide with a velocity of 1 m/sec at the top. At the bottom of the slide, she is moving horizontally, y=1.5 meters above the water. She splashes into the water d=2 meters to the left of the bottom of the slide. Assuming potential energy to be zero at the water level, what is the mechanical energy of the child at the top of the slide?
MEo=

Relevant Equations

U = mgy
K = 1/2mv^2
E = U + K

Vf = ?
y = ?

ME = mgy + 1/2mv^2
ME = 56*9.81*y + 1/2*56*1^2

Ui + Ki = Ui + Ki
gyi + 1/2vi^2 = gyf + 1/2 vf^2gyf = 1/2vf^2
vf = 5.425 m/s

9.81y + 1/2*1^2 = 9.81*1.5 + 1/2*5.425^2
y = 2.949 m

MEi = 56*9.81*2.949 + 1/2*56*1^2
MEi = 1648 J

The picture for this problem really confuses me. I am not sure what information is relevant and what is not. Do I need to break this problem into x and y components? Can I assume that the child's speed is 0 when she hits the water?

 

[haruspex]

Can I assume that the child's speed is 0 when she hits the water?

Of course not! How could that be true?

(Ah, you meant after she hits the water. That is not relevant to any equation you may write.)

You are told that she leaves the slide traveling horizontally, 1.5m above the water. What can you deduce from those two facts?

You are also told how far she travels horizontally before hitting the water. What does that tell you about her horizontal velocity?

 

[etotheipi]

Can I assume that the child's speed is 0 when she hits the water?

The problem becomes a lot more complicated once she hits the water; not least because you now have to account for all of the forces that the water exerts. It will cause her to decelerate, but not go to zero immediately.

But that's not important for the question posed here. You're only interested in the motion up to just before she hits the water. Since the normal contact force of the slide is orthogonal to the velocity it does no work, so you can safely apply conservation of energy during the motion up to just before she hits the water.

 

[Merlin3189]

Homework Statement:: A 56 kg child slides down a water slide with a velocity of 1 m/sec at the top. At the bottom of the slide, she is moving horizontally, y=1.5 meters above the water. She splashes into the water d=2 meters to the left of the bottom of the slide. Assuming potential energy to be zero at the water level, what is the mechanical energy of the child at the top of the slide?
MEo=
Relevant Equations:: U = mgy
K = 1/2mv^2
E = U + K

View attachment 265219
Vf = ?
y = ?

ME = mgy + 1/2mv^2 ME at all points = PE at bottom of slide + KE at top of slide (?)
ME = 56*9.81*y + 1/2*56*1^2

Ui + Ki = Ui + Ki PE + KE = PE + KE
gyi + 1/2vi^2 = gyf + 1/2 vf^2 I don't know what gyi and gyf are? g and y are constant. gyf = 1/2vf^2 PE at bottom = KE at bottom (?)
vf = 5.425 m/s You calc this, then reverse the calculation on the next line.

9.81y + 1/2*1^2 = 9.81*1.5 + 1/2*5.425^2 9.81y ? + KE top = 9.81*1.5 ? + KE bottom (or PE at bottom)
y = 2.949 m

MEi = 56*9.81*2.949 + 1/2*56*1^2 initial ME = PE at 2.949 m + KE at top
MEi = 1648 J

*The picture for this problem really confuses me. I am not sure what information is relevant and what is not.
*Do I need to break this problem into x and y components?
*Can I assume that the child's speed is 0 when she hits the water?

*Depends how you choose to solve the problem. Not all required, but use what you need to find the results YOU choose to calculate.
*For the ME you don't need to break the problem into x,y components, since energy is a scalar, not a vector.
But you will want to consider both horizontal and vertical motion after she leaves the slide. This is to calculate the horizontal velocity, from the given horizontal distance and height above the water as she flies off the end of the slide.
*No. It isn't (as Haruspex said.) Effectively it is this terminal velocity that you are going to calculate.

If we assume that there is no friction, air resistance, etc. then the ME does not change. There is a change of PE into KE.
Now you don't know her starting height, so PE is difficult there
When she hits the water, you are told that PE is zero, so all her ME is KE. If you happen to know how fast she is going when she hits the water, then you can calculate her KE and you've solved the problem.

So maybe forget all this energy for the moment and just think of her flying off the end of the slide like a stone thrown off the top of a cliff (or whatever examples you've done before.)

 

[Thea Woods]

Thank you all so much! I was able to for h using kinematic equations and found it to be 0.6156 m. Anyway thanks :)

 

[haruspex]

Thank you all so much! I was able to for h using kinematic equations and found it to be 0.6156 m. Anyway thanks :)

I think that's a bit inaccurate. To improve accuracy, work entirely symbolically, only plugging in numbers right at the end. This is good style always, and has numerous advantages.
You should end up with a very simple formula only involving d, h and y.
g, it turns out, is irrelevant. If you cannot get that, please post your working.