Question

A shopping bag is hanging straight down from your hand as you walk across a horizontal road at constant velocity.

Does this force do any work if you are standing in a running lift? Explain your answer and consider various frames of reference.

Answer 1

Diagram of the given problem:

Description:

1. When the lift is going up, the force from the arm is also in the upward direction, to counter the force on the shopping bag due to gravitational force.
2. By definition, work done is $W=F.S=FS\cos \left(\theta \right)$where F is the applied force, S is the displacement, and$\theta$ is the angle between force and displacement.
3. Displacement of the body is also in the same direction so $\theta =0^o$
4. Hence, $W=FS\cos \left(0\right)=F\times S$. so the work done is positive.
5. When the lift starts to come down, the force and displacement are both in the opposite directions, so $\theta ={180}^o$
6. Therefore, $W=FS\cos \left(180\right)=-F\times S$. So, the work done is negative.

Final Answer: When going up, the work done is positive. When the lift descends it's negative.