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Help, been stuck for two months

  1. Nov 24, 2003 #1
    Hey, since I am doing distance learing, my assignments don't have a date they have to be in, however I have been stuck on two questions for aBOUT 2 months now.


    1) A 0.25 kg soccer ball is rolling at 6.0m/s toward a player. The player kicks the ball back in the oppoiste direction and gives it a velocity of -14 m/s. What is the average force between the players force during the interaction between the players foot and the ball if the interaction lasts 2.0 X 10^-2 seconds.


    2)A person weighing 490 N stands on a scale in an elevator.
    - The elevator slows down at -2.2 m/s as it reaches the desired floor, what does the scale read?
     
    Last edited: Nov 24, 2003
  2. jcsd
  3. Nov 24, 2003 #2

    HallsofIvy

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    You've been stuck for two months? Exactly what have you been doing for those two months? Just staring at the problem?

    Show us what you have tried, what you understand about the problem, and where you have trouble.
     
  4. Nov 24, 2003 #3
    well I have been flipping through my books looking to find what its talking about.
    for the first one this is all I can come up with

    ------> 6 m/s
    <-------------- 14 m/s

    therefore

    <-------- 8 m/s

    the second one, I have no idea.
     
  5. Nov 24, 2003 #4

    Doc Al

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    Staff: Mentor

    For the first problem, read up on impulse and change in momentum.

    The second problem is not properly worded. If that -2.2 is meant to be an acceleration, not a speed, then it's time to review Newton's 2nd law. Ask yourself, what are all the forces acting on the person?
     
  6. Nov 24, 2003 #5

    jcsd

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    1) determine the accelaration the ball has undergone and use the formula F = ma

    2) What is the the sum of the forces acting vertically through the person?
     
  7. Nov 24, 2003 #6
    for the first one I get 909.09 is that even close to being right??
     
  8. Nov 25, 2003 #7
    Re-did the question after seeing a mistake I made and came up with the answer of -9.09. I don't believe that this is the right answer, what am I doing wrong.

    I am using the formula:

    F= ma = m * change in velocity / time.

    Is this right????
     
  9. Nov 25, 2003 #8

    NateTG

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    Try doing things one step at a time:

    For question 1:

    What is the change in velocity?

    How long does it take?

    What is the average acceleration during that time?

    What is the mass of the ball?

    What is the average force?
     
  10. Nov 25, 2003 #9
    That's right. But your answer isn't.

    Your 1st error:
    The change in velocity is not -8 m/s. Remember:
    change in velocity = final velocity - initial velocity
    = -14 m/s - 6 m/s

    Your 2nd error:
    2.0 X 10^-2 is not .22, as you assume. It's .02
     
    Last edited: Nov 25, 2003
  11. Nov 25, 2003 #10

    AD

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    1) A 0.25 kg soccer ball is rolling at 6.0m/s toward a player. The player kicks the ball back in the oppoiste direction and gives it a velocity of -14 m/s. What is the average force between the players force during the interaction between the players foot and the ball if the interaction lasts 2.0 X 10^-2 seconds.

    Rather than tease you with the answer like one might a dog with a packet of dog biscuits, I think I'll just give you the answer.

    You already know that force is equal to mass multiplied by an acceleration, which is a change in velocity over time.

    F = ma = m(&Delta;v/&Delta;t)

    Writing this differently,

    F = (m&Delta;v)/&Delta;t

    You may notice that the numerator can also be written &Delta;mv, a change in momentum. This is another definition of a force: a change of momentum over time. This will give you your answer.

    A change in momentum is

    &Delta;p = mv2 - mv1

    Calling the final and initial velocities v2 and v1 respectively.

    Since m is common to both terms,

    &Delta;p = m(v2 - v1)

    Now, the change in velocity is the final velocity minus the initial velocity. Calling the direction in which the ball is kicked by the player the positive direction, the change in velocity is

    v2 - v1 = 14 - (-6) = 14 + 6 = 20 ms-1

    You need to pay attention to the signs. Plugging this answer in to the equation worked out above:

    F = m&Delta;v/&Delta;t = m(20)/(1/50) = (50 * 20)m = 1000m

    A precise numerical answer is 1000 * 0.25 = 250 N.

    2)A person weighing 490 N stands on a scale in an elevator.
    - The elevator slows down at -2.2 m/s as it reaches the desired floor, what does the scale read?


    Now F = ma, so another way of thinking about acceleration is the force acting per unit mass, a = F/m. Now your weight is equal to

    W = mg

    Where g is the acceleration due to gravity at the earth's surface, this is roughly 9.8ms-2.

    Now, the force acting on the scales when the lift is moving at constant velocity is equal to the person's weight. Gravity is pulling the man down with a force of 9.8 newtons per kilogram. But there's another force acting on the man when the lift is decelerating.

    Let's assume that the lift is travelling upward and the upward is the positive direction. The lift will slow down as it reaches the floor and as a result, the reading on the scale will decrease.

    Call the force on the scale F, then

    F = W + ma

    Where W is the man's weight, m his mass and a the acceleration of the lift.

    F = W + m(-2.2) = W - 2.2m

    m is W/g, so

    F = W - 2.2(W/g) = g(W/g) - 2.2(W/g) = (W/g)(9.8 - 2.2) = (W/g)(7.7)

    This is equal to (420 * 7.7)/9.8 = 385 N
     
  12. Nov 26, 2003 #11
    ok,thanks you so very much every one
     
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