Incline with a Spring and Loop the Loop

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Homework Help Overview

The discussion revolves around two physics problems involving energy conservation: one concerning a block sliding down an incline and compressing a spring, and the other involving a block moving along a loop-the-loop track. Participants are exploring concepts of gravitational potential energy, elastic potential energy, and kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of conservation of energy principles, questioning how to relate potential and kinetic energy in both scenarios. There are attempts to clarify the role of height and the equations involved, particularly in the context of the loop-the-loop problem and the spring compression.

Discussion Status

Some participants have offered guidance on using energy conservation equations, while others are still grappling with how to set up their equations correctly. There is ongoing exploration of different interpretations and methods for solving the problems, particularly regarding the calculations for height and energy at various points.

Contextual Notes

Participants express uncertainty about the application of trigonometry in the incline problem and the assumptions regarding energy forms at different points in the loop-the-loop scenario. There are mentions of specific values and equations being debated, indicating a lack of consensus on the correct approach.

chaotixmonjuish
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A 3.0 kg block starts at rest and slides a distance d down a frictionless 32.0° incline, where it runs into a spring. The block slides an additional 24.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 439 N/m. What is the value of d?

img:http://i242.photobucket.com/albums/ff106/jtdla/prob01a.gif

I'm not sure how to start this. I'm not sure how to calculate potential/kinetic energy (which was suppose to be the theme of the homework). I'm aware that this problem uses Hooke's law, but I'm not sure how to put it all together.


A small block of mass m = 2.0 kg slides, without friction, along the loop-the-loop track shown. The block starts from the point P a distance h = 53.0 m above the bottom of the loop of radius R = 20.0 m. What is the kinetic energy of the mass at the point A on the loop?

img:http://i242.photobucket.com/albums/ff106/jtdla/prob17a.gif

Outside of the potential energy, I'm not sure how kinetic energy works in a circle. Or it its no different. I'm not sure if this assumption is correct, shouldn't the PE at the top of the track equal the KE at the bottom of the loop. Would that KE then equal the PE for the look itself. I'm not sure at all how to do these problems. This conservation of energy has been really tricky for me.
 
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In the first problem, use gravitational potential energy and elastic potential energy... at the top it has only gravitational potential energy... at the bottom it has only elastic potential energy... use conservation of energy.

In the second problem, it doesn't matter that it is a circle or a loop... just use conservation of energy... use gravitational potential energy and kinetic energy... again look at the beginning and at the end... at the beginning it has only gravitational potential energy... at the end it has gravitational potential energy and kinetic energy (there's no need to think about the bottom of the loop).
 
mgh= 1/2mv^2

how would I calculate the height for potential energy?

for the second problem:

I thought it would only have kinetic energy at the end. Would I just plug everything into a mgh+0=0+1/2mv^2
 
Would an equation for the loop the loop problem be
2*9.8*53=1/2*2*v^2=32.2304

or would it equal PE=KE+PE(at top of loop)
 
chaotixmonjuish said:
Would an equation for the loop the loop problem be
2*9.8*53=1/2*2*v^2=32.2304

or would it equal PE=KE+PE(at top of loop)

yes, because it has both KE and PE at the top of the loop...
 
chaotixmonjuish said:
mgh= 1/2mv^2

how would I calculate the height for potential energy?

do you mean mgh = (1/2)kx^2? use trig for height. distance = d + 0.24
so what is the height?
 
For the loop the loop, would the velocity be 25.432 m/s, would that also be the kinetic energy at A.
 
chaotixmonjuish said:
For the loop the loop, would the velocity be 25.432 m/s, would that also be the kinetic energy at A.

how'd you get that?
 
I set up that equation

PE=KE+PE1038.8(Start)=1/2*2*v^2+392(point A) v=25.4323
 
Last edited:
  • #10
Height at the top of the loop is 2*20 = 40m
 
  • #11
15.9625, I forgot about 2r

Is that the answer to the loop the loop
 
  • #12
254.801 for the kinetic energy
 
  • #13
chaotixmonjuish said:
254.801 for the kinetic energy

yeah, looks right.
 
  • #14
okay, the spring one is really throwing me off

so the distance traveled is d+.24, how can I use trig
 
  • #15
chaotixmonjuish said:
okay, the spring one is really throwing me off

so the distance traveled is d+.24, how can I use trig

you have a right triangle... angle 32. hypoteneuse d+0.24. What's the height?
 
  • #16
could i set something up like cos^-1(x/x+.24)? I had that idea earlier but I wasn't sure if that would even be close.

If that's right, i get a height of 2.536
 
  • #17
height = (d+0.24)sin32

so now just use the energy equations...
 
  • #18
Okay, how about calculating velocity. Or would I instead use the potential energy for a spring.
 
  • #19
chaotixmonjuish said:
Okay, how about calculating velocity. Or would I instead use the potential energy for a spring.

You don't need velocity. velocity at the bottom is just 0 ie no kinetic energy. just use mgh = (1/2)kx^2

solve for d.
 
  • #20
So my equation should look like

3*9.8*(sin(32)*(d+.24))=1/2*439*.24
 
  • #21
chaotixmonjuish said:
So my equation should look like

3*9.8*(sin(32)*(d+.24))=1/2*439*.24

yes, but it should be (.24)^2, for 1/2 kx^2
 
  • #22
Okay, so that yielded 57.1 cm
 
  • #23
chaotixmonjuish said:
Okay, so that yielded 57.1 cm

Looks right to me.
 
  • #24
So the simple pendulum question has a second part which I have been working on.

What is the least value that v0 must have if the cord is to swing up to a horizontal position?

Would I just solve for v0 if the angle were 90?
 
  • #25
chaotixmonjuish said:
So the simple pendulum question has a second part which I have been working on.

What is the least value that v0 must have if the cord is to swing up to a horizontal position?

Would I just solve for v0 if the angle were 90?

what is v0?
 
  • #26
We are suppose to solve for the smalles v0
 
  • #27
chaotixmonjuish said:
We are suppose to solve for the smalles v0

ok, I just looked at the problem... I think the angle of v0 is supposed to remain as 24 degrees...

use conservation of energy again... what is the initial energy... what is the final energy...
 
  • #28
What do you mean by that, just set up an equation using the same final velocity.

I've been pondering two methods, one in which neither velocity was known, but that would lead nowhere. The other was just using the same veocity at its lowest point.
 
  • #29
chaotixmonjuish said:
What do you mean by that, just set up an equation using the same final velocity.

I've been pondering two methods, one in which neither velocity was known, but that would lead nowhere. The other was just using the same veocity at its lowest point.

set the final velocity to 0. you want the minimum value of v0 so that it reaches the pendulum gets to a horizontal position... if you set final velocity>0 then you aren't dealing with the minimum...

So final energy is just mgL (setting the lowest position as 0).

What is the initial energy in terms of vo?
 

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