You are in a lift moving from the 3rdfloor to the 12thfloor, having a height difference of H = 40 m. Your mass is M = 80 kg and the acceleration due to gravity is g = 10 m/s2. If the elevator moves at a constant speed without stopping, what is the total work performed by all of the forces together (the net force) acting on you?a)Mghb)-Mghc)0d)MgCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

As the final velocity after the action of all forces is 0, therefore the person has 0 net acceleration.

And, as we know that

F_net = m.a_net

_{Therefore, the net force is 0 and hence, the net work done is also 0.}

Answer

Explanation

Answer

Net work done is equal to change in its kinetic energy
KE=1/2m(vf^2-vi^2)...since vf=vi so net work done is zero

Answer

As the final velocity after the action of all forces is 0, therefore the person has 0 net acceleration.

And, as we know that

F_net = m.a_net

_{Therefore, the net force is 0 and hence, the net work done is also 0.}

Answer

Explanation:

To find the total work performed by all the forces acting on you, we need to consider the work done by the gravitational force and the work done by the normal force.

Work done by the gravitational force:
The work done by the gravitational force is given by the formula:
W = F*d*cosθ
where W is the work done, F is the force, d is the displacement, and θ is the angle between the force and the displacement.

In this case, the force acting on you is your weight, which is given by the formula:
F = m*g
where m is your mass and g is the acceleration due to gravity.

The displacement is the vertical distance traveled by the elevator, which is equal to the height difference H between the 3rd floor and the 12th floor.

Since the elevator moves at a constant speed without stopping, the net force acting on you is zero. This means that the normal force is equal in magnitude and opposite in direction to the gravitational force.

Therefore, the angle θ between the force and the displacement is 180 degrees, and cosθ is equal to -1.

Substituting the values into the formula for work done by the gravitational force, we get:
W = m*g*d*cosθ = (-m*g*H)

Work done by the normal force:
The normal force does not do any work because it is perpendicular to the displacement. Therefore, the work done by the normal force is zero.

Total work done:
The total work done by all the forces together (the net force) is the sum of the work done by the gravitational force and the work done by the normal force.

Since the work done by the normal force is zero, the total work done is equal to the work done by the gravitational force:
Total work done = Work done by gravitational force = -m*g*H = -80*10*40 = -32000 J

However, the question asks for the total work performed by all of the forces together, which means we need to consider the direction of work.

In this case, the work done by the gravitational force is negative because the force and the displacement are in opposite directions.

Therefore, the correct answer is:
c) 0

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