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The following problem is from "The Physics Classroom": Lamar raises a loaded barbell from the floor to a position above his head with outstretched arms. Determine the work done by Lamar in lifting $300 \text{ kg}$ to a height of 0.90 m above the ground. The answer is $$2.6\times10^3 \text J$$ The audio solution says things I am confused by. First, they said Lamar's force has to equal the weight ($mg$) of the object. But why isn't the object allowed to accelerate? Secondly, because the forces are equal and opposite, the barbell must have constant velocity. But what's stopping me from saying the object has a constant velocity of zero which would mean the barbell would never move? Thirdly, $$W=\vec F\cdot\vec d$$ only applies if $F$ is constant. But if the barbell stops moving at $0.90 \text{ m}$, Lamar's force had to have switched directions from up to down to decelerate the barbell to a complete stop with the help of weight ($mg$). So if we define $F$ to be Lamar's force, $F$ is not constant, so we can't use $W=\vec F\cdot\vec d$ to compute the work done by Lamar's force.

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First, they said Lamar's force has to equal the weight (mg) of the object. But why isn't the object allowed to accelerate?

Actually, in order for Lamar to lift the barbell to get it started moving he had to briefly exert a force greater than $mg$ to initially accelerate the barbell. But before reaching 0.9 m he has to exert a force slightly less than $mg$ of an equal amount for the same amount of time to bring the barbell to rest at 0.9 m.

Secondly, because the forces are equal and opposite, the barbell must have constant velocity. But what's stopping me from saying the object has a constant velocity of zero which would mean the barbell would never move?

Because once Lamar got the barbell moving by briefly accelerating it he can reduce his force to equal $mg$ so that the barbell would continue moving at constant velocity equal to that produced by the initial acceleration. Remember that something can have constant velocity with no net force acting upon it.

Thirdly, W=Fd only applies if F is constant. But if the barbell stops moving at 0.90 m, Lamar's force had to have switched directions from up to down to decelerate the barbell to a complete stop with the help of weight (mg).

Yes he had to switch directions of the force. But the force $F$ doesn't have to be constant for $W=Fd$. $F$ can be the average force over the distance $d$. For Lamar to get the barbell moving he had to exert a force slightly greater than $mg$ at the beginning for a brief amount of time. But before getting to 0.9 m Lamar had to reduce his force to be slightly less than $mg$ by an equal amount for the same amount of time to bring the barbell to rest at 0.9 m. In that way his average force would equal $mg$.

Lamar's force had to have switched directions from up to down to decelerate the barbell to a complete stop with the help of weight (mg). So if we define F to be Lamar's force, F is not constant, so we can't use W=Fd to compute the work done by Lamar's force.

That's correct that Lamar had to switch the directions of his force from being slightly greater than $mg$ at the start to slightly less than $mg$ at the end. But, as I already indicated, the force he applied does not have to be constant in order for the barbell to start and end at rest, as long as his average force over the distance is $mg$,

So if we define F to be Lamar's force, F is not constant, so we can't use W=Fd to compute the work done by Lamar's force.

Again, $F$ does not have to be constant. Only the average value of $F$ has to be equal to $mg$ in order to apply $W=Fd$.

Hope this helps.

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  • $\begingroup$ Thank you for taking the time to answer my question. Why does Lamar's force in the beginning have to be greater than mg? My thinking is when the barbell is at rest on the ground, there is both mg acting downwards and a normal force acting upwards. So, as long as Lamar's force is greater than zero, the normal force + Lamar's force > mg so for a split second while the normal force still exists (once the barbell is off the ground the normal force disappears) the net force is upwards so the barbell will accelerate and Lamar's force can be any value greater than zero not just greater than mg. $\endgroup$ – user532874 May 30 '19 at 8:18
  • $\begingroup$ Secondly, when I said that Lamar switches the direction of his force, I didn't mean the net force but actually his force alone. My thinking is that to get the barbell to initially accelerate and then to move at a constant velocity upwards, Lamar's force is upwards. But, to decelerate the barbell so it is at rest at 0.90 m, he no longer pulls up but pulls down on the barbell. So, his force literally switches direction and with the help of mg, the barbell decelerates to zero quickly. $\endgroup$ – user532874 May 30 '19 at 8:22
  • $\begingroup$ Thirdly, how do I know that besides the one period of acceleration and the one period of deceleration, the rest of the motion is constant velocity? Why can't this "rest of the motion" also be comprised of periods of acceleration and deceleration? $\endgroup$ – user532874 May 30 '19 at 8:32
  • $\begingroup$ @user542874 Imagine the barbell on a scale. The scale reads $mg$. You grab the barbell and very gradually pull up on it watching the scale reading go down until it reads exactly zero but the barbell is not yet in motion. The net force on the barbell is now zero. To give it an upward velocity it must undergo an acceleration requiring an upward force >$mg$ no matter how small and how briefly to get it started. $\endgroup$ – Bob D May 30 '19 at 8:54
  • $\begingroup$ @user542874 Secondly you don’t pull down on the barbell. You lessen your upward force to <$mg$ so tha the net force is down. The barbell is still moving up but it is now decelerating $\endgroup$ – Bob D May 30 '19 at 9:07
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Note the Work-Energy Theorem, which gives that

the work done by all forces acting on a particle equals the change in the particle's kinetic energy + the change in the particle's potential energy.

In your case, kinetic energy is unchanged but potential energy isn't. Assuming the floor as a zero point, the change in potential energy is $mg\Delta{h}$, which is our answer, 2600J.

Finally,

But what's stopping me from saying the object has a constant velocity of zero which would mean the barbell would never move?

The fact that Lamar lifts the weight means it does not have a constant velocity of 0.

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Since the weight lifter is lifting the weights normal to the ground there is no actual work being built up .. to justify :

Work done = Force * displacement * cos(theta) work done = Force * displacement * 0 . . . . . . . . because theta here is 90 and cos 90 = 0

hence no work is done :)

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  • $\begingroup$ It's not clear why you say the angle between the force and the displacement is 90°. If you lift up a barbell, you apply vertical force and get vertical displacement. So $\cos 0° = 1$. $\endgroup$ – JMac Mar 30 at 12:54
  • $\begingroup$ Sorry but this answer must be corrected as noted already in the comments, as it stands it is just incorrect. $\endgroup$ – ohneVal Mar 30 at 13:35
  • $\begingroup$ I have voted to delete this answer due to incorrect content. $\endgroup$ – Aaron Stevens Mar 30 at 15:09

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