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Homework Help: Concept of work done

  1. Sep 30, 2009 #1
    An object is moving with an initial velocity 5m/s, if i'm applying a force against it and it stops after a displacement.

    So the work done is equal to 0.5m(0)2-0.5m(5)2

    A stationary object, 0m/s, if i'm applying a force to it and it moves with a velocity of 5m/s, and stop after the force removed.

    So the work done is equal to zero? 0.5m(0)2-0.5m(0)2=0
    Should it be zero? and why it is zero? the object did travel a distance isn't?

    definition: Work is a measure of change of energy.
    So does it mean no matter how far an object moved, or how fast it has moved, we only consider the initial velocity and final velocity?

    for eg, an object moved from stationary, speed up to 2m/s for 30min, then slower down to 1m/s and speed up to 5m/s for 1 hour, then finally it stop and remain stationary again. the work done in this case is also considered as zero?

    if i apply a force on a stationary object, and it move a certain time after the force has removed. should i take the zero as initial velocity or the instantaneous velocity when the force just applied on it?

    Please help, i'm very confused. thank you
    Last edited: Sep 30, 2009
  2. jcsd
  3. Sep 30, 2009 #2
    I think you are confused over the idea of work done on the body, and work done by the forces.
    Let's say I apply a constant force on a body to accelerate it from rest to a speed v over a smooth surface. Then, assuming that no resistive forces exist, work done by the force on the body is equal to the total net work done on the body, 0.5mv2. Then I remove this force, and it continues moving in a straight line with velocity v. For it to come to a stop, a resistive force must act on it. So, we let the body move over a rough surface. Thus, the frictional force does negative work on the body (since it is opposite in direction to the body's displacement) - i.e. extracting energy from the body. When the body comes to a rest, then you are right in saying that the net work done on the body is zero. The work done by each force on the body is however not zero, and total net work done on the body is the sum of the work done on the body by each force.
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