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Difficulty on some distance problemsneed help

  1. Aug 2, 2004 #1
    Recently took a test and got one of the lowest scores I've ever gotten....theres no tutor avaliable and the prof isnt avaliable either....I'm trying to figure out the correct answers and how and which equations to use for each problem. I always have difficulty interpreting the questions and understanding them completely....and end up using incorrect equations and so forth....These are the questions I got completely wrong on the test and need help with.....

    1. A ball is hit w/ a horizontal vector component of 15.0 m/s to the right and the vertical vector component of 20.0 m/s upward. What are these component velocities after 2.0s?
    I first found the velocities for the horizontal and vertical using v= d/t....I got this problem wrong.


    2. A car moving at 30.0 meters per second stops in 10.0 meters. How much time (in seconds) was required for the stop? Extra- What was the 'deceleration' as a factor of "g"?
    I used t=d/v and got .333s and used that to find the acceleration and also got this problem wrong.

    What tension (ie Force) must a 2.0 meter length of cord provide in order to whirl an attached 15.0 kg mass in a circular path with a tangential speed of 5.0 m/s? and can you perform this operation?

    I used Fc= mv^2/r
    = (15.0 kg)(5.0 m/s)^2
    ____________________
    3.14

    I am not sure how to tackle this problem.


    A 50.0 kg object is lifted using 4900.0 J. How high is it lifted? and what maximum kinetic energy is possible if it falls?

    I left this one blank..i was clueless

    THX
     
  2. jcsd
  3. Aug 2, 2004 #2

    Tom Mattson

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    You aren't supposed to use v=d/t for the vertical component. Gravity acts in the vertical direction, so you have vy=vyi-gt.

    You aren't supposed to use t=d/v for this problem either. You never use it when acceleration is involved. You are supposed to use:

    vf2=vi2+2a(xf-xi)

    You were given all the information called for by this equation except a, which you are supposed to solve for.

    You've got the right equation. The tension is equal to Fc. But why are you dividing by 3.14? As the equation says, you are supposed to use the radius of the circle, which is 2.0 meters.

    The gravitational potential energy (U) of the mass (m) after it is lifted to a height (h) above the ground is given by U=mgh. The maximum kinetic energy is equal to this amount, due to the law of conservation of energy.
     
  4. Aug 2, 2004 #3
    Remember that you can analyse the two components completely seperately. Remember not to get confused and consider all necessary factors (acceleration(a), time(t), initial velocity(u), final velocity(v) and distance(s).
    First consider horizontal motion:
    U = 15.0m/s. A = 0m/s, V = ?m/s, t=2.
    V = u + at

    edit: Try to finish it from there.

    Now consider vertical motion:
    U = 20, V = ?, a = -9.8 (gravity), t = 2.
    V = u + at

    edit: Try to finish it from there.

    Again, consider the factors. Don't get confused. Just list what you know, calmly and methodically:
    U = 30m/s, V = 0, S = 10m, T = ?, a = ?.
    First find the time:
    S = (u+v)/2 multiplied by T
    10 = 15T
    T = 10/15
    T = 2/3 seconds.
    Now you want to find the deceleration, which is simply a negative acceleration - ie, slowing down.
    V = u + at

    edit: Try to finish it from there.



    Not in my syllabus. Another user should be able to assist you there.


    .
    Use the potential energy equation.
    Potential energy = mgh.
    You are putting in 4900J, 50KG weight. G is gravity.

    4900 = 50g.h

    edit: Try to obtain the height from there.

    If it then falls, this gravitational potential energy of 4900J is transferred to kinetic energy.

    edit: Think about what that means for the maximum kinetic energy, given that the law of conservation of energy holds.

    The following equations may be useful to you, use them when the acceleration is considered even and not a function of time:

    For initial velocity(u), final velocity(v), acceleration(a), time(t) and distance(s):

    V^2 = U^2 + 2as
    V = u + at
    S = ut +0.5a(t^2)
    S = 0.5(u+v)t

    Hope that helped.
     
    Last edited by a moderator: Aug 2, 2004
  5. Aug 2, 2004 #4
    You were given all the information called for by this equation except a, which you are supposed to solve for.

    I dont know what xf and xi are used for, which variables?

    You've got the right equation. The tension is equal to Fc. But why are you dividing by 3.14? As the equation says, you are supposed to use the radius of the circle, which is 2.0 meters.

    I got 190 kg*m/ s^2


    The gravitational potential energy (U) of the mass (m) after it is lifted to a height (h) above the ground is given by U=mgh. The maximum kinetic energy is equal to this amount, due to the law of conservation of energy.[/QUOTE]

    to find the height i used h= pe/mg

    4900.0 j/ (50.0kg)(9.81) = h
     
  6. Aug 2, 2004 #5

    Tom Mattson

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    Gaz031--I've edited your post. Please don't post complete solutions.
     
  7. Aug 2, 2004 #6
    Thanks. I appologise.
     
  8. Aug 2, 2004 #7

    Tom Mattson

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    They are the initial and final positions of the car. Let the car start at the origin of a coordinate system (so that xi=0). In that case, xf is simply the distance traveled by the car during the braking.

    OK, now just note that 1 kg*m/s2 equals 1 N.

    Right.
     
  9. Aug 2, 2004 #8
    I think that posting answers along w/ the worked out problems helps a lot more in understanding it thoroughly....but I guess if those are the rules thats cool...Thanks to both of you I was able to understand the problems more and get the final answers.. :)
     
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