A weightlifter holds a $1,300 $N barbell $1$ meter above the ground. One end of a $2$-meter-long chain hangs from the center of the barbell. The chain has a total weight of $600$ N. How much work (in J) is required to lift the barbell to a height of $2$ m?
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?
For number one, I am aware that the formula for this problem is Work = Force x Distance. However, I don't know how this is supposed to be set up. I have previously tried to add the two forces ($1300$ N and $600$ N) together and multiplied it by the distance ($2$ m), but it was wrong. I'm not sure what else could be done here.
As for number two, I am not sure where to go from there, either. All I know is that I have to set up an integral from 0 to 2 with respect to dy and multiply that value by a scale factor of 1/2.
Update: The chain that has not been lifted initially is assumed to still be on the ground. So the higher the barbell is lifted, the more chain that is being lifted from the ground.
Let's assume the weight$-$lifter is located at a point $2$ meters above the ground, is pulling the barbell up to his position, and can allow the part of the chain he has already lifted to go slack.
With this interpretation, we see that, initially, there is $1$ meter of chain hanging with the $1300N$ barbell attached at the other end; the weight$-$lifter need only lift $1$ meter of the chain up to his position along with the $1300N$ barbell.
Consider when the weight$-$lifter has lifted $y$ meters of the chain up to his position, where $0\leq y\leq 1$.
This means there is $1-y$ meters of chain left hanging which weighs $300(1-y)N$ along with the $1300N$ barbell.
The amount of work required to lift the chain an additional $dy$ meters is around $$dW=[300(1-y)+1300]dy$$
Therefore $$W=\int_{y=0}^{y=1}[300(1-y)+1300]dy=1450$$
is the total amount of work required.