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Homework Help: Conservation of Energy Supp. 2

  1. Mar 18, 2004 #1
    A 2kg block is released 4m from a massless spring with a force constant k=100N/m that is along a plane inclined at 30 degree. A) if the plane id frictionless find the maximum compression for the spring. b) if the coefficient of kinetic friction between the plane and the block is 0.2, find the maximum compression. c) for the plane in part b) how far up the incline with the block travel after leaving the spring?
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  3. Mar 18, 2004 #2


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    A) The block is released 4m (measured along the slant) above the spring and the incline is at 30 degrees. What is the vertical height of the block initially above the spring? What is its potential energy relative to the spring? How far does the spring have to be compressed in order for that energy to be converted to work compressing the spring?

    B) With how much force does the block press against the plane? Since the friction coefficient is 0.2, how much is the friction force?
    Taking "x" to be the distance the spring is compressed and applying that friction force for 4+ x m, how much work is done by the friction force on the block? How much work is done by the spring in being compressed x m? Add those "works", set equal to the potential energy you got in (A) and solve.

    C) When the block "bounces" again, the work done by the spring will be cancelled by the work done by friction on the block. How far must the block move so that the work done by friction is the same as the work done by the spring in (B)?

    (If you are wondering where all that energy went, it is heat in the block, plane, and spring.)
  4. Mar 19, 2004 #3
    can you please give me a formula to start with, I am really lost with this one :(
  5. Mar 19, 2004 #4


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    It's a simple energy conversion between gravitational and elastic energy.

    [tex]\frac{1}{2}kx^2 = mgh[/tex]

    substitute in

    [tex]\frac{1}{2}(100)x^2 = (2)(9.8)(4sin(30))[/tex]

    solve it from there

    This one is like the previous one but you have to factor in the work done by friction. The two sides of the equation represent work in and work out (left side is out, right side is in).

    [tex]\frac{1}{2}kx^2 + Fd = mgh[/tex]

    expand it once

    [tex]\frac{1}{2}kx^2 + uNd = mgh[/tex]

    substitute in

    [tex]\frac{1}{2}(100)x^2 + (0.2)(2)(9.8)(cos(30))(4) = (2)(9.8)(4sin(30))[/tex]

    solve from there

    Here, you take the answer for x you got in part B and use that to solve for the other parts of the equation. This time it's different because the energy from the spring is the work put in; friction and gravitational energy are the work taken out, that means friction goes to the other side of the equation now.

    [tex]\frac{1}{2}kx^2 = uNd + mgh[/tex]

    Since the question asks how far up the slope, I think it means slant distnace.

    [tex]\frac{1}{2}(100)x^2 = (0.2)(2)(9.8)(cos(30))d + (2)(9.8)(dsin(30))[/tex]

    It's a very ugly equation but once you simplify it, you end up with a nice and easy to solve equation.
    Last edited: Mar 19, 2004
  6. Mar 19, 2004 #5


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    a) The plane is at 30 degrees to the horizontal so the block, 4m above the spring as measured along the plane is 4 sin(30)= 2 m vertically above the spring (draw a picture and look at the right triangle). The potential energy of the block relative to the spring is mgh= 2(9.81)(2)= 39.24 Joules. When the spring is completely compressed, the work it has done on the block, letting x be the amount of compression of the spring in meters, ((1/2)kx2) must be equal to that: (1/2)(100)x2= 50x2= 39.24 so x2= 39.24/50=0.7848 and x= 0.88 m or 88 cm.

    b) The component of weight perpendicular to the plane is 19.62 cos(30)= 19.62(√(3)/2)= 17 N. Since the coefficient of friction for the plane is .2, the friction force is (.2)(17)= 3.4 N.
    The total work done on the block, again letting x be the amount of compression of the spring, is (1/2)kx2+ (4+x)(3.4). (Since the block starts 4 m from the spring, the total distance it slides against friction is 4+ x.) We must have 50x2+ 3.4x+ 13.6= 39.24 or 50x2+ 3.4x- 25.64= 0. That's a quadratic equation and can be solve by the quadratic formula:
    [tex]x= \frac{-3.4+\sqrt{3.4^2-4(50)(-25.64)}}{2(50)}[/tex]
    [tex]x= \frac{-3.4+\sqrt{5139.56}}{100}=[/tex]= 0.68 m or 68 centimeters.
    (That's a "reasonable" answer. Since we now have friction also retarding motion, this should be slightly less than the answer in (a).)

    c) The work done in compressing the spring to that 68 centimeters is (1/2)(100)(0.68)= 34 Joules. Once the spring is "released" that energy will go into kinetic energy until all the kinetic energy has either been converted to potential energy (the blocks height) or has been canceled by the work done on the block by friction. If the block moves a distance x up the plane, its height increases by x sin(30)= (1/2)x, so its potential energy increases by (2)(9.81)(0.5x)= 9.81x and the work done by friction is 3.4x. We must have 9.81x+ 3.4x= = 13.21x= 34. Well, that's easy: the block will move 34/13.21= 2.57 m which is 2.57- .68= 1.89 m. above the equilibrium point for the spring and 4- 1.89= 3.11 meters below the original position of the block.
    (Again, that's a "reasonable" answer. If there were no friction at all, we would expect the block to return to its original position. Friction reduces its motion.)
    Last edited by a moderator: Mar 19, 2004
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