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Potential energy and work by spring.

  • #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!
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,428
1,824
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
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.
 
Last edited:
  • #4
ehild
Homework Helper
15,428
1,824
So the total energy at B is 3693 J, what is the velocity there?

ehild
 
  • #5
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
ehild
Homework Helper
15,428
1,824
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
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?
 
Last edited:
  • #8
ehild
Homework Helper
15,428
1,824
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
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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