# Energy skier Problems

1. Feb 20, 2009

### zell_D

Hey guys, so after dealing with forces we started learning energy related problems. And I do get some problems but things get some what complicated. There are 3 questions that I need help on, the last one being the most difficult. One of them is conceptual that i wish to understand.

1. The problem statement, all variables and given/known data

1. a 55 kg skier skis down a smooth frictionless ramp, slows down over a distance of 25 m on the flat portion at the bottom as shown in the diagram for number 1.
a) The skier pushes off at the top with a initial speed of 10 m/2. Find her speed at the top of the middle hump
b) The skier comes to rest at the end of the rough section. Find the average friction force experienced by the skier

2. You release a frictionless cart at the top of each of the 2 ramps. On ramp B, the cart is released from rest while the Ramp A cart is pushed with some initial velocity. The ramps have the same height as each other. The only difference is that one has a small bump up and the other a bump down. Which cart has the larger speed at the finish? why?

3. a) calculate the velocity of the ball when it just leaves the launching ramp (both magnitude and direction. be sure to specify a coord system.)
b) spring constant k = 800.0 N/m, the spring's compression is 3.00 cm, the mass of the ball is 50.0g, the height of the ramp is 15.0, and H (table's height) is 1.00 meter. With what total speed will the ball hit the floor? (use g = 10 m/s^2)

2. Relevant equations
1. a) 0=$$\Delta$$KE+$$\Delta$$PE
b) Work done by friction = $$\Delta$$KE+$$\Delta$$PE

2. 0=$$\Delta$$KE+$$\Delta$$PE

3. 0=$$\Delta$$KE+$$\Delta$$PE+$$\Delta$$PEspring

3. The attempt at a solution
1. a) Since no friction, I used 0=$$\Delta$$KE+$$\Delta$$PE which will become=
0=[1/2(mv^2)f-1/2(m[10^2])]+[mg(hf-hi)]
50m=1/2(mv^2)f+[mg(hf-hi)]
50=1/2(v^2)f+[9.8(0-20)]
100=(v^2)f+[-196]
296=v^2f
v=17.2 m/s

b) Since friction, I used Work done by friction = $$\Delta$$KE+$$\Delta$$PE
and since I obtained the velocity from part a, I assume I can start off with the velocity I obtained as the initial? (Confused as to whether or not I can do this?)

Work done by friction = $$\Delta$$KE+$$\Delta$$PE
Ffriction(Cos(180))$$\Delta$$x=[1/2(m0^2)f-1/2(m[17.2^2])]+[mg(0-10)]
0=-1/2(m[17.2^2])]+[mg(-10)]-Ffriction(cos(180))25.0
1/2(m[17.2^2])]+[mg(10)]=Ffriction(25.0)
Ffriction=541 Joules

2. Since no friction, I used 0=$$\Delta$$KE+$$\Delta$$PE again
change in PE will both be the same since the change in height was the same, the bumps does not matter since the difference between the heights are still the same. so this also implies that the change in kinetic energy will also be the same. This then means that the change in KE will be the same between Ramp A and B. However, since Ramp A is pushed with some initial velocity, it starts off with some KE, will be going that much faster because of it.

on second thought, for 3a, these values such as the spring constant and mass wasn't given. They were only given in part 3b. Should I assume I can use them? because if not, i have no idea how to do this problem.
3. a) E1 (energy at spring, only spring potential)=1/2(k)x^2 = 1/2(800)(.03^2) = .36 J
E2 (top of ramp, both KE and PE)=mgh + 1/2(mv^2) = .05(10)(.15)+1/2(.05)v^2
Conservation of energy due to no friction
E1=E2
.36=.075+.025v^2
.285/.025=v^2
v=3.38 m/s @ top of ramp
however, how do i calculate the direction of this?

b) NOT sure
E3 introduced: energy of ball on the floor = mgh + 1/2(mv^2)
E3=E1=E2
.36=.050(10)(-1.0) + 1/2(.050)v^2
.86 = .025v^2
34.4 = v^2
v = 5.87 m/s as it hits the floor?
[negative 1.0 meters for table length because the height of the table that is set to be 0]

some of these i probably did wrong. But i would like a check and some advice is possible. thx.
Thank you and sorry for the long post. I read the book but some concepts are just hard to grasp and visualize

Last edited: Feb 20, 2009
2. Feb 20, 2009

### LowlyPion

For the first question part a) looks ok, though I think it may be conceptually a little easier thinking about it as you start out with a KE, you add some gravitational PE and that equals your final KE.

Part b) Consider the total KE at the bottom. You start with an initial KE you add the PE of gravity all the way down - 30m - and that is your kinetic energy at the bottom. What slows that down? it will be Work. And Work is Force over a distance. So no need to calculate the velocity at the bottom. Just figure the force directly by W = F*d = the KE at the bottom = Ke at top + Gravitational PE.

#2 is correct.

3. Feb 20, 2009

### LowlyPion

For question 3) I think you are right. I think they want a general formula since no values are given until part b).

So for part b) what is the angle θ? Isn't that given?

4. Feb 20, 2009

### zell_D

also found that i did my arithmetic wrong on 1a haha

for your help on 1b, what do i assume as initial velocity then? cuz i need that to figure out the initial kinetic energy at the top right?

3a okay so general formula wise, what would be the direction of the final velocity? would it be Vx? since the ball would be propelled forward by the spring horizontally?

3b the angle isn't given

Last edited: Feb 20, 2009
5. Feb 20, 2009

### LowlyPion

If all they want is the |V| you know that from the conservation of energy.

m*g*H + 1/2kx2 = 1/2*m*V2

6. Feb 20, 2009

### zell_D

which part are you replying for? for 3a they want to know the magnitude and direction

for 3b they want to know speed

if thats for 3b, can you explain why wouldnt the equation be
1/2kx2 - m*g*H = 1/2*m*V2 instead?

since PE(spring)+KE(initial) = KE(final) + PE(final) ? or am i totally wrong

I don't know what the initial ke is here though...

Last edited: Feb 20, 2009
7. Feb 20, 2009

### LowlyPion

I was only talking about b).

When the spring is compressed it has +PE stored. That becomes KE when released. Then add to that the m*g*H. The rise in h is not relevant once the ball falls below the horizontal as it is a conservative field.

8. Feb 20, 2009

### zell_D

okay so let me get this straight
I am okay on the concept of the elastic potential energy
my question is, when the spring is not FULLY compressed, will there be a KE? or can i just assume the velocity is zero and therefore no KE

you said that the PE spring becomes KE when released, so where was this taken into account? Sorry for the bombardment of questions>.<

and for 3a, the velocity i find is the overall velocity, do i need to break it down into components or what? cuz i dont know the direction of it since it goes up the ramp right?

9. Feb 20, 2009

### LowlyPion

For b) note at some point before it is released it is not in motion.
The total energy then is the Potential in the spring and the potential by being H off the ground.
That's all there is. No kinetic. No rotational. Just those 2.

When it's released lots of things happen, but one thing remains the same - total energy.

When it strikes the floor what's available? Spring PE long gone. Gravity PE just hitting 0. So where did it all go? Must be in kinetic energy then.

For part a) you can always say Vo = VoCosθ i + VoSinθ j, where |Vo| = x*(k/m)1/2

10. Feb 20, 2009

### zell_D

ok i think i see it.

now back to part a i guess. i don't see what the angle has go to do with anything. sure i know the velocity and can use pythagorean (sp) to find the x y components... but how does that help for direction

11. Feb 20, 2009

### LowlyPion

The only way to describe the Vo in a coordinate system and give a direction is by using the angle θ.

Otherwise it's simply

|Vo| = x*(k/m)1/2 at an angle θ to the horizontal.

12. Feb 20, 2009

### zell_D

hmmm, how would you do it if say the numbers from b worked for a? (problem 3)

also, for problem 1b, can i still assume the initial velocity is 10 m/s ?

13. Feb 20, 2009

### LowlyPion

Just plug in the numbers.

Since they say the initial speed in 1 is 10m/s why would you think not?