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Average force on a baseball

  1. Mar 8, 2005 #1
    I need help with this question as well: If the baseball in example 4.5 is caught by a person whose hand recoils 0.30 m, calculate the average force on the person's hand. (The initial speed of the baseball is 44.7 m/sec and its mass is 0.50 kg.)

    This is what i have, but i'm not sure if it's right...

    w=mg i'm not even sure if i have to convert that or not
    w=(0.50 kg)(9.8 m/sec2)
    w=4.9N


    then i'm not sure how to represent kenitic energy on here but here it goes,
    KE=(1/2mv)2 ..(squared)
    KE= 1/2(4.9 N)(44.7 m/sec)2
    KE= 4895.32 J

    thats what i have, but im not sure if i should convert mass to weight if not then go i just put the 0.50 in the KE equation? Any help would be appreciated.
     
    Last edited by a moderator: Sep 9, 2015
  2. jcsd
  3. Mar 8, 2005 #2
    Use Newton's 2nd Law: [tex]a = \frac{F_{net}}{m}[/tex] or [tex]F_{net}=ma[/tex]

    To find "a" is the tricky part. You have to use the information they gave you. I would use the kinematic equation:

    [tex]v_{f}^2 = v_{i}^2 + 2ad[/tex]

    Vf = 0
    Vi = 44.7
    d = .3
    a = ?

    You should be able to solve it from here.
     
  4. Mar 8, 2005 #3
    Thanks =)
    ok for acceleration i got -73.5 m/sec

    then i did
    F=ma
    F=(0.50 kg)(-74.5 m/sec)
    F=37.25 N

    now my question is can you have negative force, cause i got a negative number ..but i didnt think you could have negative force.
     
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