constant speed = normal force & weight are the same ?

Question

A wiehgt lifter can lift 150kg vertically at a constant speed to a position 2m over his head.

so because he is lifting it at a constant speed, when you draw the free body diagram, you know that the normal force and the weight should be the same (due to equilibrium). So does that mean if you get the sum of the work done by the weight and the normal force, it should be ZERO?

weight is gravity times the mass ,
so i am not forgetting the gravity :)

Answer 1

> if you get the sum of the work done by the weight and the normal force, it should be ZERO?

That's exactly right.

The work done by the weighlifter (i.e., by the normal force) is +2940 Joules. This work is positive because the force is in the same direction as the motion.

The work done by gravity is −2940 Joules. This work is negative because the force is in the _opposite_ direction from the motion. (Force down; motion up).

The total work done (by all sources) is the sum of these, which is zero.

This is consistent with the work-energy principle; which says that the total work done on an object should match its change in kinetic energy. Since this object's kinetic energy doesn't change (i.e., it continues to move at a constant speed), it follows that the net work done on the object is zero.

Answer 2

The normal force is a perpendicular force from the contact surface. Now, what contact surface is this normal force arising from? The force form the weightlifter is the applied force. Once the weights are off the ground, the normal force from the ground is no longer acting on the weights. Thus, it does no work. The forces acting on the weights once it has left the ground are the force of gravity (its weight) and the force from the weightlifter. The work done by gravity is -mhg. The work done by the weightlifter is mhg. The net work is zero when the weights are brought to a stop (work-kinetic energy theorem)

Answer 3

that all depends on the time he's taking to do it. he should be moving at about 3 meters for every 4 minutes of accuracy . 150+2=152. 152/4= 38 so you have 38 meters of accuracy so far. in order to reach an equilibrium, he'd have to increase the speed by 5 more minutes of accuracy.

Answer 4

commonly used rigidity exerted via the seat is the reaction rigidity (equivalent and opposite to)to a mix of the downwards rigidity of gravity plus the different forces performing. At relax there is merely gravity. on the backside of the wheel there is the greater rigidity by using rotation of the wheel performing downwards to be further to gravitational rigidity- so commonly used rigidity is larger than while at relax. on the actual of the wheel there is greater rigidity by using rotation performing upwards and so this is subtracted from the rigidity of gravity so the conventional rigidity would be below while at relax. so the suitable commonly used rigidity is on the backside of the wheel

Answer 5

Your entire question is a bunch of mis-conceptions.
Additonal details reveal more mis-conseptions.
Such is the sad state of no child left behind education.

The force of gravity acting on the barbel is always the same, regardless of accleration: F = mg. This force acts on the barbel.

Weigth of the babel W is another force. This force acts not on the barbell. Weight of the barbel W acts on the hands of the lifter.

According to third law of Newton this force is accompanied by its re-action counterpart force. This is force N is applied by the lifter to the barbell, and according to the third law of Newton this force N is ALWAYS equal in magnitude and opposite in direction to weight W of the barbell:
N = -W.
These two forces are ALWAYS equal regardless of acceleration of the barbell.

Now comes the second law of Newton. The two forces acting on the barbell are:
* force of gravity mg
* reaction N
Note that weight of the barbel W is not on the list. Wieight of the barbel acts not the barbell, it acts on the lifter.
2nd law of Newton
ma = mg + N
0 = mg + N
N = -mg.
In case there's no accleration reaction N must be equal to force of gravity mg. Which in turn means (because N = -W at all times) that
W = mg.

Conclusion:
despite weight W and force of gravity mg act on different bodies, their magnitudes and direction are eqaul in case the barbell does not accelerate.


Now:
The work done by the athlete is
(force applied by athlete to the barbel) x (distance)
Work done by the athlete = N * ∆h = -mg ∆h
This work is non-zero, and positive, as one would expect from common sense. Weight lifting is hard work.


The work done ON the barbell by ALL forces is
(force applied by athlete to the barbel) x (distance) +
+ (force of gravity acting on the barbel) x (distance) =
= (N + mg) ∆h = 0 * ∆h = 0
The work done ON the barbell by ALL forces is ZERO.
Because of this its kinetic energy of the barbell does not change.

Answer 6

You are forgetting gravity.