# Incline with a Spring and Loop the Loop

#### chaotixmonjuish

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 [Broken]

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 [Broken]

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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#### learningphysics

Homework Helper
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).

#### chaotixmonjuish

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

#### chaotixmonjuish

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)

#### learningphysics

Homework Helper
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...

#### learningphysics

Homework Helper
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?

#### chaotixmonjuish

For the loop the loop, would the velocity be 25.432 m/s, would that also be the kinetic energy at A.

#### learningphysics

Homework Helper
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?

#### chaotixmonjuish

I set up that equation

PE=KE+PE

1038.8(Start)=1/2*2*v^2+392(point A) v=25.4323

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#### learningphysics

Homework Helper
Height at the top of the loop is 2*20 = 40m

#### chaotixmonjuish

Is that the answer to the loop the loop

#### chaotixmonjuish

254.801 for the kinetic energy

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#### chaotixmonjuish

okay, the spring one is really throwing me off

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

#### learningphysics

Homework Helper
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?

#### chaotixmonjuish

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

#### learningphysics

Homework Helper
height = (d+0.24)sin32

so now just use the energy equations...

#### chaotixmonjuish

Okay, how about calculating velocity. Or would I instead use the potential energy for a spring.

#### learningphysics

Homework Helper
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.

#### chaotixmonjuish

So my equation should look like

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

#### learningphysics

Homework Helper
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

#### chaotixmonjuish

Okay, so that yielded 57.1 cm

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#### chaotixmonjuish

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?

#### learningphysics

Homework Helper
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?

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