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F=ma and SUVAT

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    "A metal ball of mass 0.50 kg is dropped from the top of a vertical cliff of height 90m. When it hits the beach below it penetrates at a depth of 6.0cm. Calculate:
    a) the velocity of the ball just as it hits the sand
    b) the (average) retarding force of the sand

    2. Relevant equations

    SUVAT equation: v2=u2+2as
    Newton's Second Law: F=ma

    3. The attempt at a solution

    For part a) I used suvat (s=90m, u=0ms-1, v=v ,a=9.81ms-2 , t=t) and used the suvat equation to calculate the final velocity as 42ms-1

    For part b) I am unsure, do I just use F=ma (with m=0.50kg and a=9.81ms-2) to get an answer of 4.91N, or am I missing something?
     
  2. jcsd
  3. Nov 16, 2013 #2

    cepheid

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    For part b, no, it's not just the case that a = g. That would be true only if the object were in free fall (meaning gravity is the only force acting). Once it hits the ground and starts burrowing, this is no longer true. There is another force (from the ground).

    What you have to do is apply the same equation as you did for part a, this time with u = the v you calculated from part a, and v = 0 since it comes to rest. From this you can figure out what acceleration is *required* in order to slow the object down over that distance. This required acceleration tells you the net force.
     
  4. Nov 17, 2013 #3
    Part a) is alright .

    For Part b) the ball starts with an initial velocity 42ms-1 and ends up being at rest covering a distance of 6.0 cm .During the motion the ball is under the influence of two forces,force of gravity and the retarding force of the sand .Using F=Ma with a=9.81 is incorrect. Value of 'a' can be calculated using SUVAT equations .Now using F=Ma will give you the net force .
     
    Last edited: Nov 17, 2013
  5. Nov 17, 2013 #4

    CWatters

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    You should find that a > g due to the very short stopping distance.
     
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