1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force needed to jump 0.8 m

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data
    An exceptional vertical jump from rest would raise a person 0.8m off the ground. To do this, what constant force would a 70kg person have to excert against the ground? Assume that the person lowers himself by 0.2m prior to jumping and remains in a standing position while in the air. The answer is 3400N.


    2. Relevant equations
    aΔt=v2-v1
    d= 1/2(v1 + v2)∆t
    d=v1Δt + 1/2aΔt^2
    v2^2=v1^1 + 2ad
    F=ma
    μ=Ff/Fn


    3. The attempt at a solution

    I know acceleration must be found by finding the velocity of the person while in the air and while on the ground, just taking off. I do not know how to go about finding the two velocities . I would then put them in the fourth formula as it omits time
     
  2. jcsd
  3. Nov 5, 2012 #2

    I like Serena

    User Avatar
    Homework Helper

    Welcome to PF, Monic! :smile:

    The easiest way to solve this problem is not with speeds but with conservation of energy.
    The energy generated is distance times the unknown force.
    When at the highest point, the energy consists solely of potential energy, which is mass times the acceleration of gravity times the height.

    Are you aware of those relationships?
     
  4. Nov 5, 2012 #3
    After the person leaves the ground, only gravity will be acting on him. So he must leave the ground with some initial velocity to reach a height of 0.8m. You should be able to calculate that initial velocity and use that information to complete the problem.
     
  5. Nov 5, 2012 #4
    I am aware of conservation of energy and tried it but it did not seem to work, I must be making an error.

    I do not know how to find the initial velocity... would v^2=2(9.8)(0.8m) be correct? v=3.96m/s as v?


    I am really lost on this one :( thank you for the welcome :)
     
  6. Nov 5, 2012 #5
    For the conservation of energy technique: force * distance = change in potential energy

    Be careful with the change in potential energy side of the equation. That is probably where you are making your error.

    For velocity/acceleration/gravity approach you could determine the time to rise (or fall) 0.8 m by gravity. Knowing that time, you can easily compute what the initial velocity is. Knowing that, use v2^2=v1^2 + 2ad to determine acceleration. Follow with Newton's second law for the average force.
     
  7. Nov 5, 2012 #6
    At the risk of bumping into ilikeserena,

    Actually yes that works -- I hope you understand why :) The person will be leaving the ground with initial velocity 3.96 m/s.

    I thought in terms of energy conservation too, so just so you understand how it is done, I'll do this part for you to show you come to the same answer.

    State 1: person leaves the ground with unknown initial velocity v
    State 2: person reaches maximum height at 0.8m where velocity is zero

    Energy state 1 = Energy state 2
    0.5mv2 + mg(0) = 0.5m02 + mgh
    v2 = 2gh

    which was your equation. I chose the ground as height zero in calculating gravitational potential energy.

    Next you need to find what constant force the person applies through 0.2m to achieve that initial speed. There is a difference between the net force the person experiences and the total force he needs to apply.


    ILikeSerena wanted you to do it all in one step. You may want to see if you could apply conservation of energy to do that.
     
  8. Nov 5, 2012 #7
    okay so the initial velocity is 3.96 m/s and v2 is 0m/s at the top of the jump, therefore

    0^2=3.96m/s^2 +2(a)0.8

    giving a= 9.8....which is wrong

    FInding the constant force that needs to be applied at 0.2m to get this person that high must be greater than the normal force f=mg which is 686N in this case. I do not think any of my equations will give me that force
     
  9. Nov 5, 2012 #8

    I like Serena

    User Avatar
    Homework Helper

    Seeing you're getting splendid help with the velocity approach, I'll restrict myself to the conservation of energy approach. ;)

    What did you try?

    You're welcome. :wink:
     
  10. Nov 5, 2012 #9
    We split the problem into two parts. Part 1 has the person applying a force through 0.2m before leaving the ground. Part 2 has the person leaving the ground and achieving a height of 0.8m.

    We solved part 2 to find that the person leaves the ground at 3.96 m/s.

    We now have enough information to solve part 1. The person starts at rest, applies a constant force through 0.2m to achieve a speed of 3.96 m/s. A little investigation will reveal all forces acting on the person are constant which means the person will experience constant acceleration. You should be able to use the same formula as you did before v2 = 2ad. (Keep in mind this formula assumes a constant acceleration which is why I kept emphasizing that). The force applied is ma. But you are not done because this is the net force required to move the person. Are there other forces the person must overcome to achieve this net force?

    A conservation energy approach for just this part1 has an outside force acting through a distance of 0.2m. The person starts at rest and achieves 3.96 m/s. Gravity is also involved.

    This problem is certainly easier to solve using conservation of energy all in one step and does not depend on assumptions of constant acceleration. The tricky part of the one-off conservation equation is detailed in the last paragraph. Once you've solved it the velocity way, you should take a quick look at the conservation method.
     
  11. Nov 5, 2012 #10
    okay all I did was

    v^2=2gh
    v=2(9.8)(0.8)
    v=3.95

    the other velocity is 0

    then I put 0^2=3.95^2 +2a(0.2)
    to find the acceleration needed to find the force for f=ma
    a=39

    f=39(70kg)
    f=2730N which is still wrong, unless added the normal force he needs to overcome would be the correct thing to do, (686N)

    That seems to get the answer 3400N, but the method seems dubious to me
     
  12. Nov 5, 2012 #11
    This is right, assuming the math is right. But this is the net force needed to generate the person's acceleration.

    It is :) Imagine yourself standing. Gravity is pulling down on you. If you didn't apply an opposite force with your muscles, you would fall down. The act of not moving in the presence of gravity requires your body to apply an equal and opposing force. If you want to move in addition to that, you will have to apply force above and beyond that.

    Free body diagrams help to clarify these things. In the case of the person, we are talking about the motion of his center of gravity (for rigid bodies it turns out a body moves as if all external forces are applied at the center of gravity).

    Represent the person as a point at his center of gravity. Standing still has the free body diagram looking like:

    Flegs
    ^
    |
    o
    |
    v
    Fgravity

    The net force on the point is Flegs-Fgravity. If that is zero, it doesn't move.

    If it is not zero, the point will accelarate upward according to Newton's Law:

    Fnet = (Flegs - Fgravity) = ma

    When you found a and multiplied by mass you found the net force.
     
    Last edited: Nov 5, 2012
  13. Nov 5, 2012 #12

    I like Serena

    User Avatar
    Homework Helper

    If you want to do the problem with conservation of energy, try to do it without any velocities at all.
    Just the initial state at speed zero and the highest state that also has speed zero.
     
  14. Nov 5, 2012 #13
    Finally it all makes sense!!! :D thank you a lot to everyone, as for conservation of energy I just don't know how to apply it to this problem as it uses joules not force but thats fine
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force needed to jump 0.8 m
  1. Force required to jump (Replies: 7)

  2. Force when jumping (Replies: 4)

Loading...