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Work and Distance help!

  1. Sep 26, 2006 #1
    I have a quick question, I need to find Work for this problem, but I don't have a distance!? The problem provides vi and vf, time, and mass, but no distance? How do i find work?

    1. A 1.50 X 10^3 kg car accelerates from rest to 10m/s in 3 seconds.
    A. What is the work done in this time? W=?
    B. What is the power delivered by the engine? P=?

    W= Force * Distance (cos theta) <-- Right?

    W=? P=? m=1500 kg vi=0m/s vf=10m/s T=3 seconds

    -------------------------------------------------------------
    ******************Edit**************************
    -------------------------------------------------------------

    I solved my own problem!
    x = .5 (vi+vf)time

    Anyone wanna check my work for me?

    W=? P=? m=1500 kg vi=0m/s vf=10m/s T=3 seconds x=?

    x= .5 (0+10) 3
    x= 15m

    Force = mass * acceleration
    F = 1500kg * 10...?
    F = 15000 J?

    W = 15000 * 15
    W = 225000 J...?

    Thats alot of work!?

    -------------------------------------------------------------

    After somebody could nicely check my work i have one more problem...

    A skateboard reaches a speed of 35m/s from an initial speed of
    25m/s after 21KJ of work

    I THINK what the problem is asking for here is: What is the skateboarder's kinetic energy disregarding the mass of the skateboard?
     
  2. jcsd
  3. Sep 26, 2006 #2

    mezarashi

    User Avatar
    Homework Helper

    Unfortunately, I don't think your approach was correct. Although, you would have arrived at the same answer (given you had used the correct value for acceleration under the assumption that it is constant), I think it is better you use concepts from Work & Energy rather than mixing in Kinematics unnecessarily.

    Firstly, the equation: [d = 0.5(Vf+Vi) x t] only works when the acceleration is constant. There is no reason to assume this from what is given.

    Going back to the first principles of work and energy: the work done on an object over a period of time or distance is equal to the change in kinetic energy of that object.

    From the question, can you calculate the kinetic energy (0.5mv^2) of the car before the acceleration? And after? What is the difference? By conservation of energy, how much energy should have been delivered?

    Power is Energy/time. If that given energy above was delivered in 3 seconds, what is the power?
     
    Last edited: Sep 27, 2006
  4. Sep 27, 2006 #3
    thats simple
    according to ur data,
    vi=0 vf=10m/s and t=3secs
    u can find acceleration simply by using acceleration formula
    a=(vf-vi)/t => a= 10/3 => a= 3.33m/(sec^2)
    now putting it in F=m*a where m=mass
    F=m*a
    now u can find force
    about distance u can apply
    d= (vi*t) + (a*(t^2))/2
    as Force and distance both are in hand
    W= F*d and W can be calculated easily
    after that P=W/t
    so power is also calculated
    hope u get the answer correct.........
     
  5. Sep 27, 2006 #4
    Ok, lets re-do this:

    vi=0 vf=10m/s t=3secs m=1500 kg

    a=(vf-vi)/t
    a=(10)/3
    a=3.3m/s^2

    d= (vi*t) + (a*(t^2))/2
    d= (0*3) + (3.3(9))/2
    d= 14.85m

    F = m * a
    F = 1500 * 3.3
    F = 4950

    W = F * d
    W = 4950 * 14.85
    W = 73507.5 <--- ?
     
  6. Sep 28, 2006 #5
    I'd love it for someone to check my latest work :!!)
     
  7. Sep 28, 2006 #6
    Well
    ur work is gr8
     
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