homework and exercises - Work done by a weight lifter

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.

Answer

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.