Potential energy and work by spring.

In summary, a 15.0 kg stone slides down a snow-covered hill, 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.
  • #1

Homework Statement



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/problemAsset/1260235/2/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?

Homework Equations



PE = mgh
KE = .5mv^2

spring = .5kx^2

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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  • #2
Crusaderking1 said:
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
 
  • #3
ehild said:
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.
 
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  • #4
So the total energy at B is 3693 J, what is the velocity there?

ehild
 
  • #5
ehild said:
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.
 
  • #6
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
 
  • #7
ehild said:
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?
 
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  • #8
Crusaderking1 said:
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
 
  • #9
ehild said:
it looks all right, but the red 6 has to be 7.


ehild

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

1. What is potential energy?

Potential energy is the stored energy that an object possesses due to its position or state. It is the energy that can be converted into other forms of energy, such as kinetic energy, when the object is in motion.

2. How is potential energy related to a spring?

A spring has potential energy when it is stretched or compressed from its equilibrium position. The amount of potential energy stored in a spring is directly proportional to the amount it is stretched or compressed.

3. What is the equation for calculating potential energy of a spring?

The equation for potential energy of a spring is PE = 1/2kx², where k is the spring constant and x is the displacement from equilibrium. This equation only applies to ideal springs that follow Hooke's Law.

4. How does work relate to potential energy of a spring?

Work is done on a spring when it is stretched or compressed, causing it to store potential energy. This work is equal to the change in potential energy of the spring, as described by the work-energy theorem.

5. Can potential energy be negative?

Yes, potential energy can be negative. This occurs when the object is in a position or state that is lower than its equilibrium position, such as a compressed spring. Negative potential energy represents the work that would need to be done to bring the object to its equilibrium position.

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