# Potential energy and work by spring.

Tags: energy, potential, spring, work
 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!
HW Helper
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P: 10,770
 Quote by Crusaderking1 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
P: 159
 Quote by ehild 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.

 HW Helper Thanks P: 10,770 Potential energy and work by spring. So the total energy at B is 3693 J, what is the velocity there? ehild
P: 159
 Quote by ehild 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.
 HW Helper Thanks P: 10,770 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
P: 159
 Quote by ehild 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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P: 10,770
 Quote by Crusaderking1 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
P: 159
 Quote by ehild 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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