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Potential energy and work by spring. 
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#1
Oct3011, 12:08 AM

P: 159

1. The problem statement, all variables and given/known data
A 15.0 kg stone slides down a snowcovered 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/YF0734.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! 


#2
Oct3011, 02:31 AM

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


#3
Oct3011, 11:25 AM

P: 159

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. 


#4
Oct3011, 11:30 AM

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P: 10,615

Potential energy and work by spring.
So the total energy at B is 3693 J, what is the velocity there?
ehild 


#5
Oct3011, 11:38 AM

P: 159

In that case, velocity would equal 22.2 m/s. 


#6
Oct3011, 11:55 AM

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P: 10,615

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
Oct3011, 11:59 AM

P: 159

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? 


#8
Oct3011, 05:46 PM

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P: 10,615

ehild 


#9
Oct3011, 06:20 PM

P: 159




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