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Average force in stopping a block

  1. Jul 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A 300kg block is suspended by a cable.
    a) what is the weight of the motor block?
    b) the cable snaps and the block falls 3 metres to the ground below.
    i) calculate the time taken for the block to reach the ground.
    ii) calculate the impact velocity of the block with the ground
    c) if the block comes to rest having caused a 1cm dent in the block (ie travelled 1cm in coming to rest), calculate the average force exerted by the ground in stopping the block.

    2. Relevant equations
    a) weight = mass x gravity
    b) i) s = ut + 0.5at^2
    ii) v = u + at
    c) ?


    3. The attempt at a solution
    a) weight = 300x9.81 = 2943N down
    b) i) 3 = 0.5 x 9.81 x t^2 t = 0.78s
    ii) v = 9.81 x 0.78 = 7.65m/s down
    c) no idea
     
  2. jcsd
  3. Jul 17, 2009 #2

    rock.freak667

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    Work done against the block = change in gravitational potential energy of the block.
    (Conservation of energy)
     
  4. Jul 17, 2009 #3
    work done = potential energy
    w = m.g.h
    w = 300x9.81x3
    w = 8829 joules

    is that right?
     
  5. Jul 17, 2009 #4

    rock.freak667

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    Right, so if you have the work done on the block. How would you find the force if you are given the distance?
     
  6. Jul 17, 2009 #5
    work = force x displacement

    foce = 8829/3

    force = 2943N
     
  7. Jul 17, 2009 #6

    rock.freak667

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    The block travels 1cm before coming to rest.
     
  8. Jul 17, 2009 #7
    2943 x 0.01?
     
  9. Jul 17, 2009 #8
    the unit of 2943 x 0.01 is not Newton (the question is about force)

    rock.freak667 talked about what the force is if the block travelled 1 cm with some amount of work done.
     
  10. Jul 18, 2009 #9
    so it's 2943x0.01 J?
     
  11. Jul 18, 2009 #10
    What is the unit of force?
     
  12. Jul 18, 2009 #11
    Newtons
     
  13. Jul 18, 2009 #12
    will the unit of this calculation be Newton?
     
  14. Jul 18, 2009 #13
    Newton seconds?
     
  15. Jul 18, 2009 #14
    lol

    yes it's Newton seconds and that's unit of momentum, not force.
    Since the question is about force, no doubt the answer is wrong.

    you have the distance and work, find F
     
  16. Jul 18, 2009 #15
    s = 0.01m
    W = 300x9.81x3
    F = W/s
    F = 882900N
     
  17. Jul 18, 2009 #16

    Redbelly98

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    Looks good.

    I recommend getting in the habit of including units during your calculations, and not just sticking them on at the end. Including units should help you understand better what's going on, and will help you catch errors like thinking that "2943 x 0.01" represents a force.
     
  18. Jul 18, 2009 #17
    Ok, thanks for the advice! :biggrin:
     
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