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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


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    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)
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