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Potential energy and work by spring.

by Crusaderking1
Tags: energy, potential, spring, work
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Crusaderking1
#1
Oct30-11, 12:08 AM
P: 159
1. The problem statement, all variables and given/known data

A 15.0 kg stone slides down a snow-covered hill (the figure ), leaving point A with a speed of 10.0 m/s. There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100m and then runs into a very long, light spring with force constant 2.50 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

Here is the picture: http://session.masteringphysics.com/...2/YF-07-34.jpg

A) What is the speed of the stone when it reaches point B?

B) How far will the stone compress the spring?



2. Relevant equations

PE = mgh
KE = .5mv^2

spring = .5kx^2


3. The attempt at a solution

A) I did this:

PE = (15.0)(9.80)(20.0) = 2943 J
KE= .5(15.0)(10.0)^2 = 750 J

2943+750 = 3693 J

3693 = .5(15.0)(v)^2-.5(15.0)(10.0)^2
v=19.80 m/s

B) work done by spring = .5(2.50)(x)^2 = 1.25x^2
work done by friction = (0.2)(15.0)(9.8)x = 29.4x

Using quadratic:

1.25x^2 + 29.43x - 3693 = 0

x= 43.84 or -67.38
x= only 43.84 meters.

Do these answers seem right? thanks!
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ehild
#2
Oct30-11, 02:31 AM
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P: 10,347
Quote Quote by Crusaderking1 View Post
A) What is the speed of the stone when it reaches point B?

A) I did this:

PE = (15.0)(9.80)(20.0) = 2943 J
KE= .5(15.0)(10.0) = 750 J

2943+750 = 3693 J

3693 = .5(15.0)(v)^2-.5(15.0)(10.0)
v=19.80 m/s
You miss a square it the KE term: It has to be .5(15.0)(10.0)2

I do not understand what you mean with the second term when you calculate the velocity at B.


ehild
Crusaderking1
#3
Oct30-11, 11:25 AM
P: 159
Quote Quote by ehild View Post
You miss a square it the KE term: It has to be .5(15.0)(10.0)2

I do not understand what you mean with the second term when you calculate the velocity at B.


ehild
yes, thanks. It should be .5(15.0)(10.0)^2 = 750 J.

Wouldn't the total energy = KE + PE, so 750 Joules+ 2943 joules be the total work at point B? Then set 3693 joules equal to .5mv^2-.5mvo^2, which game me 19.80 m/s.

ehild
#4
Oct30-11, 11:30 AM
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P: 10,347
Potential energy and work by spring.

So the total energy at B is 3693 J, what is the velocity there?

ehild
Crusaderking1
#5
Oct30-11, 11:38 AM
P: 159
Quote Quote by ehild View Post
So the total energy at B is 3693 J, what is the velocity there?

ehild
Oh ok. I think that I only need to set the joules equal to the final velocity, since the initial is 0.

In that case, velocity would equal 22.2 m/s.
ehild
#6
Oct30-11, 11:55 AM
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P: 10,347
Correct!
Now, question B. Before the stone reaches the spring, it slides 100 m on the rough ground. So it has lost a lot of energy when reaching the spring.

ehild
Crusaderking1
#7
Oct30-11, 11:59 AM
P: 159
Quote Quote by ehild View Post
Correct!
Now, question B. Before the stone reaches the spring, it slides 100 m on the rough ground. So it has lost a lot of energy when reaching the spring.

ehild
thanks! I will try figure part B out now.

ok, I added work done by friction. I received (15.0)(9.8)(0.2)(100)cos180 = -2940 joules.

-2940 + 3693 joules = 653 joules.

Would the new quadratic equation be 1.25d^2 + 29.4d -653 = 0?
ehild
#8
Oct30-11, 05:46 PM
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P: 10,347
Quote Quote by Crusaderking1 View Post
t

-2940 + 3693 joules = 653 joules.

Would the new quadratic equation be 1.25d^2 + 29.4d -653 = 0?
it looks all right, but the red 6 has to be 7.


ehild
Crusaderking1
#9
Oct30-11, 06:20 PM
P: 159
Quote Quote by ehild View Post
it looks all right, but the red 6 has to be 7.


ehild
ah ok. good catch. Thanks a lot for the help!


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