Calculating Work Done on Moving Object: Initial vs Final Velocity

In summary, the work done in this situation is equal to zero since the net work done on the body is zero. This is because the initial and final velocities are the same, so the change in energy is also zero. However, this does not mean that the work done by each individual force is zero. The net work done is the sum of the work done by all forces on the body.
  • #1
MechaMZ
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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
 
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  • #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.
 
  • #3


In both scenarios, the work done is equal to zero. This is because work is defined as the product of force and displacement, and in both cases, the displacement is zero. The object is not moving in the first scenario, and in the second scenario, it returns to its initial position after the force is removed.

It is important to note that work is not just a measure of distance traveled, but also takes into account the force applied to cause that movement. In the first scenario, the force applied was not enough to cause any displacement, so no work was done. In the second scenario, the force applied caused the object to move, but then it returned to its initial position, so the net work done is still zero.

In your example of an object moving at different speeds and stopping, the work done would still be zero because the displacement is zero in the end. It does not matter how fast or far the object moved, what matters is the net displacement.

As for your last question, the initial velocity should be taken as the velocity at the moment the force is applied. This is because work is being done to cause the movement, so we want to consider the initial conditions when the force is first applied.

I hope this helps clarify any confusion. Work can be a tricky concept to understand, but remember that it is a measure of both force and displacement.
 

What is the formula for calculating work done on a moving object with initial and final velocity?

The formula for calculating work done on a moving object with initial and final velocity is W = (1/2)mvf2 - (1/2)mvi2, where W represents work, m represents mass, vf represents final velocity, and vi represents initial velocity.

What is the unit of measurement for work?

The unit of measurement for work is joules (J).

Can the work done on a moving object be negative?

Yes, the work done on a moving object can be negative. This occurs when the force applied to the object is in the opposite direction of its motion, resulting in a decrease in its kinetic energy and a decrease in work done.

How does the work done on a moving object change with an increase in initial or final velocity?

If the initial or final velocity of a moving object increases, the amount of work done on the object will also increase. This is because the object has a greater kinetic energy, which is directly proportional to its velocity.

Is the work done on a moving object affected by its mass?

Yes, the work done on a moving object is affected by its mass. The greater the mass of the object, the more work is required to change its velocity. This is reflected in the work formula, where mass is a factor in calculating work done.

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