Solved:YAPhysicsT: Skier going off a ramp (mechanical energy)

[Ricemarine]

- A 60 kg skier leaves a ski ramp going 24 m/s
- The ramp is 25 degrees above the horizontal
- The skier lands 14 m below the ramp @ 22 m/s

How much of the mechanical energy (Earth System) is reduced because of air drag?

So

K1 + U1 = K2 + U2 + E(thermal)

K1 = 0.5(60kg)(24m/s)^2
K2 = (0.5)(60kg)(22m/s)^2
U2 = 0 (since the y in mgy is 0)
U1 = 0 or (60kg * 9.81m/s^2 * 14m)???

Is the height negligible in this equation, thus making the potential initial energy (U1) 0 since all of it is in kinetic energy? Or is somehow the degree of the ramp and height used in this problem?


Thanks in advance.

Edit: Solved this a while ago a few posts down after Fleshlight's response... The problem I had was that it was hard to judge without the initial height of the start of the ramp, but that was cleared up rather quickly.

 

[Howard]

Yeah, the skier falls down from the ramp so he has gravitational potential energy.

 

[I Saw OJ]

When did OT become a place to get your homework done for you?

 

[FleshLight]

U1 = mgh = 60 * 9.81 * 14

You don't need toe angle of the ramp unless you need to calculate acceleration or horizontal distance.

 

[magomago]

Just depends how you set your frame of reference...and for the record i thought U was internal energy till I realized it meant kinetic energy in this case

 

[TecHNooB]

Originally posted by: I Saw OJ
When did OT become a place to get your homework done for you?

It's a Saturday night. Clearly, there is some intent by the op on actually doing the assignment..

 

[FallenHero]

the potential energy in this is my fist punching you in the face for making my brain hurt.

 

[Ricemarine]

Originally posted by: FleshLight
U1 = mgh = 60 * 9.81 * 14

You don't need toe angle of the ramp unless you need to calculate acceleration or horizontal distance.

So it should be...

17280 + 8240 - 14520 = E(thermal)

E(thermal) = 11000 J.

Does it seem about right?... Because 11000J seems kind of high. But I looked at other ski problems and it seems ok, but I just want to make sure.
Or is it that the reason for the angle is due to the increase in potential energy?

 

[Howard]

Originally posted by: FallenHero
the potential energy in this is my fist punching you in the face for making my brain hurt.

Sorry, that's kinetic energy.

 

[Howard]

11 kJ is not that high.

 

[OdiN]

How fast is the treadmill going?

 

[imported_LoKe]

Originally posted by: I Saw OJ
When did OT become a place to get your homework done for you?

At least he tried to solve it himself before posting.

 

[91TTZ]

Originally posted by: Howard

Originally posted by: FallenHero
the potential energy in this is my fist punching you in the face for making my brain hurt.

Sorry, that's kinetic energy.

But there is potential for more beating after the initial strike.

 

[marvdmartian]

Sorry, can't answer your question using that fancy equation stuff you posted, but can answer it with a common sense answer.

If the skier is still above the ground when he leaves the end of the ski ramp, and there's any height he can fall down to, he still has potential energy. Think of it like this.....what if he walked down the ski ramp, stood at the end, and jumped.....would he have potential energy the nanosecond before he jumped? YES. So when he leaves the end of the ramp, he has transferred some of his potential energy to kinetic energy, but still retains a portion of his potential energy.

If you ski down a slope, next to the ski jump, and ski to the bottom of the slope, then you have converted all your potential to kinetic energy, which will eventually dissipate due to friction losses.

I'm no physicist, but I can figger out the simple stuff, eh?

 

[DrPizza]

What do you think: "- The skier lands 14 m below..." means?? Do you think it's just there to confuse you?

 

[Eeezee]

The height tells you your initial potential energy. The angle of the ramp is unimportant in telling you that because you know that the guy is 14m above the earth, giving him initial potential energy in addition to his initial kinetic energy. You also know that he hits the ground with some final kinetic energy but no potential energy.

A better problem would have had a skier going down an incline plane, and you'd have to integrate over the distance he travels to find the total work done by friction. This problem is too easy for the end of November. But sadly, that's what's happening in the courses I'm teaching as well. Frankly, energy should be learned a lot earlier; right after kinematics and some introductions to free body diagrams and force laws should come energy, and from then on you should be able to use energy to solve almost every problem because it's so easy to use.

Here is my solution

0.5*m*v^2 + mgh = total energy (K1+U1)
0.5*m*v2^2 + Drag = total energy (K2+Ethermal)

Ethermal = 0.5*m*v^2 + mgh - 0.5*m*v2^2
Ethermal = 10992 Joules lost to drag

 

[Ricemarine]

Originally posted by: Eeezee

0.5*m*v^2 + mgh = total energy (K1+U1)
0.5*m*v2^2 + Drag = total energy (K2+Ethermal)

Ethermal = 0.5*m*v^2 + mgh - 0.5*m*v2^2
Ethermal = 10992 Joules lost to drag

Yeah, so I'm about right then.
The way my book had this planned would start with the basics of

-Units
-Velocity and Acceleration
-Vectors and the scalar product
-Average/Instantaneous velocity in 2D and 3D (projectile motion covered too)
-Newton's laws w/ the FBD's and such
-Drag force and terminal speed
-Kinetic Energy and Work
-Potential Energy
-Center of mass and linear momentum
+ the rest 2 weeks before the final...

So in a sense, there is a lot to be learned for the basics in terms of how complex you want to be about it.

I was just basically stuck on whether or not to input the potential energy, but it made sense after realizing potential energy 14m up had to be inputed.
But thanks guys for your help ( for those that did )!