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!

Particle is moved - Work-kinetic energy

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.00 kg particle is moved from the coordinates (1.00, 3.00) m to (5.00, 7.00) m.

    (q1) How much work is done by the force F = <2x^2, 0> on the particle? (Assume that the force is in Newton when x is in metres.)

    (q2) If the particle has start velocity v_i = <0, 4.00 m/s>, what is the final velocity v_f? (Hint: the "work - kinetic energy"-theorem can be useful.)

    (q3) Determine the rate of energy transfer (i.e. the effect P) from the force to the particle, at start and final position. (Answers: P_i = 0 W, P_f = 455 W.)

    2. Relevant equations



    3. The attempt at a solution

    (q1) I made a position vector AB = <4, 4>. It takes the particle from (1, 3) to (5, 7). We consider the positional vector's component that is parallell with F.

    --> x = 4 m --> W_F = 2 * 4^2 = 32, that is 32 N.


    (q2) Start velocity v_i = <0, 4.00 m/s> is constant in y-direction since the only working force on the particle is F, but this force is perpendicular on v_i, and will therefore not alter it.

    "Work - kinetic energy"-theorem:

    W_total = (delta)K = (1/2)*m(v_F)^2

    <--> v_F = sqrt( (2(W_F)) / m ) = sqrt( (2*32 N) / (2 kg) ) = 4sqrt(2) m/s

    --> v_f = <v_F, v_i>

    --> ||v_f|| = sqrt((v_F)^2 + (v_i)^2) = sqrt( (4sqrt2 m/s)^2 + (4.00 m/s)^2 ) = 4sqrt3 m/s =~ 6.9 m/s

    that is in the direction of the vector AB, 45 degrees anticlockwise with the x-axis.


    (q3) Effect from force on particle:

    at start: P = dW/dt = F[/*v_i = F * v_i * cos(alpha) = 2*0^2*cos(pi/2) = 0.
    ie 0 W

    at end: P = dW/dt = F*v_f = F*v_f*cos(theta) = 2* 4^2 * 4sqrt3 * cos(pi/4) W = 64sqrt6 W =~ 157 W.

    answer: at start: 0 W, at end: 455 W

    I must have done something wrong here. I believe the error follows from my try at (q2). Please comment or help me with getting this problem right. Thanks a lot for your time.
     
  2. jcsd
  3. Feb 6, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    For 1) I would suggest that your answer isn't in the right units for work. You got 32 N and W is in N-m.

    Specifically W = ∫ F(x)⋅dx

    Since your Force has no component in Y then you would use x2 as your F(x). You would evaluate that over the range of x from 1 to 5. This method will yield work as N-m.

    Using this result - the work supplied - and starting with an initial kinetic energy given by ½mv² after adding the work to that you can determine the velocity at (5,7)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Particle is moved - Work-kinetic energy
Loading...