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Force Problems

  1. Dec 13, 2009 #1
    I cannot solve these two problems and do not know where to start. please help. thanks.

    1. A 77 kg man drops to a concrete patio from a window only 0.45 m above the patio. He neglects to bend his knees on landing, taking 2.0 cm to stop.
    (a) What is his average acceleration from when his feet first touch the patio to when he stops?

    (b) What is the magnitude of the average stopping force?

    2. Imagine a landing craft approaching the surface of the moon of a distant planet. If the engine provides an upward force (thrust) of 2910 N, the craft descends at constant speed; if the engine provides only 2200 N, the craft accelerates downward at 0.47 m/s2.
    (a) What is the weight of the landing craft in the vicinity of moon's surface?

    (b) What is the mass of the craft?

    (c) What is the magnitude of the free-fall acceleration near the surface of the moon?
     
  2. jcsd
  3. Dec 13, 2009 #2
    A couple of hints:

    for the first problem : use the eqn: 2ax=Vf^2-Vi^2

    Note that vf = 0
    Just need to compute Vi, the speed at which he strikes the ground. X above is the displacement during deceleration, 0.02m

    Second problem: Need to understand that in the first case, he is coasting, and the rocket is providing just enough thrust to offset gravity of moon.

    F1-m*g=0 where g is the moon's gravity constant and F1 is the force of the rocket engine

    Second case there is acceleration


    F2-mg=m*a where a is the acceleration given. Pay attention to signs.

    Need to solve these simultaneously--subtracting the top from the bottom eqn seems easy.
     
    Last edited: Dec 13, 2009
  4. Dec 13, 2009 #3
    thanks but i cannot get the first problem or part b for the second one can you please explain?
     
  5. Dec 13, 2009 #4
    I'll try but you may need to spend some time looking at your book/notes as well.

    For the first: The equation to compute a, the acceleration, is given above.

    It simplifies to a(2*d)=v^2 where d is the displacement during which he stops= 2.0cm

    The same eqn can be used to compute how fast he is going when he strikes the ground:

    here 9.8(2*0.45m) = v^2

    since V^2 is common to both equations: set the other quantities as equal and solve for a.


    Problem two: If you solve the 2 simultaneous equations, you need to figure out what g is. Note both m and g are unknown, but you have 2 eqns so both can be figured out. As I hinted subtracting the first eqn from the second will get you F2-F1 = m*a and a is given to you. Now go back using that info using either equation to compute g.
     
  6. Dec 13, 2009 #5
    i got it

    thank you
     
  7. Dec 13, 2009 #6
    great, and you are welcome.
     
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