Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mechanical energy

  1. Feb 17, 2004 #1
    Hello, ive got a problem ive got stuck with. Anyone who can help me solve this?

    A 4kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6kg blovk that rests on a shelf. The coeff of kinetic friction is 0,2. The 6kg block is pushed against a spring, compressing it 30cm. The spring has a force constant of 180nm.

    How do i fins the speed och the 6kg block after its released and the 4kg block has fallen 40cm?

    Anyone out there who can help me?

    best reagards
    Johan
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Calculate the forces. The force of gravity is acting on 4 kg block. That is 4(9.81)= 39.24 Newtons downward. There is the spring acting on the 6 kg block. The spring has a force constant of 180 N/M (not "nm") so the force is 180(.3-x) where x is the distance (in meters) the 6 kg block has moved. Notice that once the 6 kg mass has moved 30 cm, the spring becomes stretch and the force is now negative (to the left. I am taking the downward force of gravity positive because that corresponds to a rightward pull on the 6 kg mass). There will also be a friction force on the 6 kg mass: its weight (6(9.81)= 58.86 N times the coefficient of friction: 0.2(58.86)= -11.77 N (negative because, as the block moves to the right, the force is back to the left).

    That is, assuming that both masses move as a unit, the force is
    39.24+ 180(.3-x)- 11.77= 39.24+180(.3)- 11.77- 180x= 81.47- 180x. The total mass of the blocks is 4+6= 10 kg. so
    ma= 10 dv/dt= 81.47- 180x or dv/dt= 8.147- 18x.

    The interesting part of this problem is that it is not obvious that we can assume that. You will need to calculate the forces on the two blocks separately (gravity 39.24 Newtons on the 4 kg block, spring and friction 42.23- 180x on the 6 kg block.
    Solve 4dv1/dt= 3924 and 6dv2/dt= 42.23- 180x separately for the speeds of the 4 kg block and 6 kg block respectively. As long as v2 is larger than v1, the top block is "catching up" to the lower block so the string has no effect. If v2 is less than v1, the string slows down the lower block and speeds up the upper block- you need to treat this as one unit as above.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Mechanical energy
  1. Mechanical energy in EM (Replies: 21)

Loading...