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Conservative and nonconservative potential energy

  1. Oct 13, 2003 #1
    I have several problems that I've been working on for homework, but it's just not really clicking with me. If anyone can help, that would be terrific. I know these are a lot of problems to throw on here at once, but I'm really frustrated and can't find anything that will work.

    1) A 110 N child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy of the child--Earth system relative to the child's lowest position at the following times. (a) When the ropes are horizontal. (b) When the ropes make a 30 degree angle with the vertical. (c) When the child is at the bottom of a circular arc.

    I tried the formula U = mgh, but I can't get it to work. For (a) I had (110 N)(9.8 m/s^2)(0). For (b) I figured out the hypotenuse of the triangle to be 4, and used that for length, and for (c) I used 2 for length.

    2) This one has a picture with it that I'll try to explain. There's a ball at the top of a slide (point a) (height = 6.00m) that's mass is 6.50 kg. The slide dips down and then curves up, and at the top of that curve is point b which is 3.20 m high. Then it dips down again, where point (c) is, which is 2.00 m high. I'm supposed to determine the ball's speed at points b and c and then determine the net work done by the force of gravity in moving the particle from A to C.

    3) A 0.408 kg ball is thrown straight up into the air and reaches a maximum altitude of 20.0 m. (a) What is its total mechanical energy? (b) What is the ratio of its kinetic energy to the potential energy of the ball-Earth system when the ball is at an altitude of 10.00 m?

    I know that the change in mechanical energy is kinetic energy plus potential energy, but I don't know how to get either of those.

    4) A parachutist of mass 59.0 kg jumps out of a balloon at a height of 1400 m and lands on the ground with a speed of 6.20 m/s. How much energy was lost in kJ to air friction during this jump?

    5) A child's pogo stick stores energy in a spring (k = 2.40 104 N/m). At position A (xA = -0.130 m), the spring compression is a maximum and the child is momentarily at rest. At position B (xB = 0), the spring is relaxed and the child is moving upward. At position C , the child is again momentarily at rest at the top of the jump. (a) Calculate the total energy of the system if both potential energies are zero at x = 0.(b) Determine xC.(c) Calculate the speed of the child at x = 0.(d) Determine the value of x for which the kinetic energy of the system is a maximum. e) Calculate the child's maximum upward speed.

    6) A block of mass m = 3.50 kg situated on a rough incline at an angle of = 37.0° is connected to a spring of negligible mass having a spring constant of 100 N/m. The pulley is frictionelss. The block is released from rest when the spring is unstretched. The block moves 15.0 cm down the incline before coming to rest. Find the coefficient of kinetic friction between block and incline.
     
  2. jcsd
  3. Oct 13, 2003 #2

    HallsofIvy

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    Technical point: there is no such thing as "non-conservative potential energy". The whole concept of potential energy exists only for conservative force fields.

    Now for your questions.
    #1 "I tried the formula U = mgh, but I can't get it to work. For (a) I had (110 N)(9.8 m/s^2)(0)."

    Your main problem is that "110 N" is not mass- it is weight which means it is gravitational force- it IS "mg". Your second problem is that in (a), h is NOT 0. The problem said specifically "relative to the child's lowest position". The child will be at the lowest position when the ropes are vertical, not horizontal. When the rope is horizontal, making a 90 the child's height is the radius of the circle: 2 m (WHERE are this child's parents? Why are they allowing him to do such a dangerous thing!). The child's potential energy is
    weight*h= 110N*2m= 220 Joules.
    For b), the "hypotenuse of the triangle" is not 4- it is the length of the swing ropes, 2m. They make a 30 degree angle with the vertical and you should be able to calculate that the height is 2 cos(30)= √(3). The potential energy is 110 √(3) Joules.

    For c) it specifically says "the child is at the bottom of the arc". The height above the bottom is 0 and the potential energy relative to the bottom is 0.

    Pretty straight forward. Since you are given heights, take height 0 as your base for potential energy. Now you can use "mgh" (notice that you are given mass: kg not Newtons) to find the potential energy (relative to height 0) at each point. The nice thing about using "conservation of energy" is that all the movement in between is irrelevant. Assuming the ball is not moving at the top, its total energy is the same as its potential energy: mgh= (6.5)(9.8)(6).
    At point b, the potential energy is mgh= (6.5)(9.8)(3.2) so the potential energy has decreased by (6.5)(9.8)(6- 3.2)= (6.5)(9.8)(2.8) Since total energy is conserved, that lost potential energy has gone into kinetic energy, (1/2)mv2 and from that you can calculate the speed.
    At point c, h= 2 m so the potential energy is (6.5)(9.8)(2) and the "lost" potential energy is (6.5)(9.8)(6-2)= (6.5)(9.9)(4) and again, that is kinetic energy so you can calculate v.
    Since gravity is a conservative force, the work done by gravity in moving from a to c IS the change in potential energy you just calculated.

    At its highest point, the ball's speed is 0 so its kinetic energy is 0. Its "total mechanical energy" is just its potential energy: again mgh= (0.408)(9.8)(20). At 10 m, calculate its potential energy: mgh= (0.408)(9.9)(10). Of course, its total energy has not changed so its kinetic energy is (0.408)(9.8)(20)- (0.408)(9.8)(10) which is obviously just (0.408)(9.8)(10) again! Now, to get the ratio, just divide- without even doing the arithmetic you should be able to see that the ratio is 10/20= 1/2.

    You know the original potential energy when he jumps: mgh= (59)(9.9)(1400) and there is no initial speed so that is his total energy. When he lands, his speed is 6.20 so his kinetic energy is (1/2)mv2= (1/2)(59)(6.20)2. Since he is on the ground, his potential energy is 0 and that kinetic energy is total energy. Subtract to find how much energy was lost.

    I may look at the other two later, but give them a try yourself first.
     
  4. Oct 13, 2003 #3
    Thanks so much .. that really helps to clarify things a lot. I'm only a little confused on two points. First, number 1 letter b. I figured the length of the leg to be 1.73 by multiplying cos 30 by 2, the length of the ropes. Then I multiplied 1.73 by 110 N to get the gravitational potential energy and got 190 J, but apparently that isn't correct. What am I doing wrong?

    Second, number 3 part c. The net work is supposed to be in terms of joules, and I was a bit confused as to how the velocity in earlier parts of the question played out into the net work. I tried to use the formula W = mgd, but was unsure of how to find the distance while missing one of the other "legs" of the triangle.

    Thanks again! I will get to those other two problems tomorrow.
     
  5. Oct 13, 2003 #4
    1b>
    In U=mgh, h is height, right? That's height ABOVE your reference point, which in this case means height above the lowest position of the swing. Look at your diagram of the swing when it's at a 30-degree angle from the vertical (I hope you drew a diagram.) Draw a horizontal line from the swing to the vertical, forming a triangle with a hypotenuse of length 2m. Now, what does 2.00*cos30 give you? It gives you the length of the long leg of that triangle, right? But is that the height of the swing? I'll let you figure that out.

    2c>Starting from rest at 6.00m, the ball has only potential energy. As it descends, some (or all, depending on height) of that potential energy becomes kinetic energy, but, if there is no friction, the total energy remains constant. If you can determine the amount of kinetic energy at a given point [Ek=mv2/2], you can figure out the velocity.
     
  6. Oct 14, 2003 #5

    HallsofIvy

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    The gravitational potential energy is 110√(3) which is approximately 190 Joules (more accurately 190.3 Joules).

    I assume you mean number 2 part c. There is no part c of number 3! You know the potential energy at the beginning- there is no kinetic energy and you can calculate the potential energy from the given height, 6 meters- the total energy is (6.5)(9.8)(6) Joules. You can calculate the potential energy at point c: (6.5)(9.8)(2) Joules. Of course, the change in potential energy is (6.5)(9.8)(6-2)= (6.5)(9.8)(4). THAT is the work done by gravity.
     
  7. Oct 14, 2003 #6
    No, for part 1b, the height is not 2cos30 (or √(3)); that is the vertical distance measuring DOWN frcom the pivot point. Instead, you want to use the height measuring UP from the lowest point of the arc.
     
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