A weightlifter jerks 220kg vertically through 1.5m and holds still at that height for 2 minutes. The work done by him in lifting and in holding it still is respectively?

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220J, 330J

3234J, 0

2334J, 10J

0, 3234J

answer is B.

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Detailed Solution

Concept- Given,
A weightlifter jerks 220kg vertically to a height of 1.5m and holds it still at that height for 2 minutes.
m=220kg
S=1.5m
g=9.8m/s
We must ascertain what he did to raise it and keep it steady for two minutes.
Let's first determine how much labor he put into lifting.
We know that work done is the product of force and displacement in the direction of force.
In equation form it can be written as
W=FScosθW
Where,
F is the force, S is the displacement and θ is the angle between the direction of force and the direction of displacement.
When lifting, we are aware that the body is going higher and that a force is at work. Therefore, the displacement is in the force's direction.
Hence,
The angle between the force and displacement is zero.
Here,
Force is equal to weight.
F=mgF=mg
⇒F=220×9.8N
On substituting all the values in the equation, we get
⇒W=220×9.8×1.5×cos0
⇒W=220×9.8×1.5×1
⇒W=3234J
This is the work done in lifting.
Now,
Let us calculate the work done in holding it for 2 minutes.
In this case there is no displacement.
⇒S=0
Since work is a function of force and displacement, it will be 0 if displacement is zero..
Hence, the correct option is 1.

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