Timeline for Work done by a weight lifter
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1, 2019 at 23:02 | vote | accept | CommunityBot | ||
| May 30, 2019 at 9:25 | comment | added | Bob D | Let us continue this discussion in chat. | |
| May 30, 2019 at 9:19 | comment | added | Bob D | @user542874 you can do all the accelerating and decelerating you want as long as the final velocity at 0.9 m is zero. By the work energy theorem the net work done on the barbell equals its change in kinetic energy. Since the initial and final velocity is zero the change in kinetic energy and net work done on the barbell is zero. You do positive work on the barbell and gravity does an equal amount of negative work taking the energy you gave it and storing it as gravitational potential energy. | |
| May 30, 2019 at 9:16 | comment | added | user225790 | I still don't get why you can't pull down on the barbell. I understand lessening the upward force to less than mg to decelerate the barbell but what's stopping Lamar from just pulling down. | |
| May 30, 2019 at 9:15 | comment | added | user225790 | Oh I see. I was thinking of the normal force and lamar's force as unrelated when in fact they are related. As lamar's force increases, the normal force decreases so the net force will continue to stay zero. Eventually, the normal force will be zero and lamar's force upwards will be equal to the weight downwards, and the net force is still zero. At this point, Lamar's force has to be greater than mg to accelerate the barbell from rest. Am I correct? | |
| May 30, 2019 at 9:07 | comment | added | Bob D | @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 | |
| May 30, 2019 at 8:54 | comment | added | Bob D | @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. | |
| May 30, 2019 at 8:32 | comment | added | user225790 | 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? | |
| May 30, 2019 at 8:24 | vote | accept | CommunityBot | ||
| May 30, 2019 at 8:24 | |||||
| May 30, 2019 at 8:22 | comment | added | user225790 | 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. | |
| May 30, 2019 at 8:18 | comment | added | user225790 | 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. | |
| May 30, 2019 at 4:13 | history | edited | Bob D | CC BY-SA 4.0 |
added 50 characters in body
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| May 30, 2019 at 4:07 | history | answered | Bob D | CC BY-SA 4.0 |