Is Holding a Weight at Arm's Length Considered Work?

[blackout85]

A boy holds a 40N weight at arm's length for 10s. His arm is 1.5m above the ground. The work done by the force of the boy on the weight he is holding. He would not be doing any work, because their is no motion and he is lifting it up.

Thank you

 

[neutrino]

He would not be doing any work, because their is no motion and he is lifting it up.

A better way to state the answer would be to say that the displacement of the object is zero. For if he had taken the object along some crazy path and brought it back to where he started, the work done would still be zero, even though the object had moved.

 

[kyle8921]

for your question. I would like to clarify some concepts related to work and provide a response to your content. Firstly, work is defined as the product of force and displacement in the direction of the force. In simpler terms, work is done when a force causes a displacement. In this scenario, the boy is indeed exerting a force of 40N on the weight he is holding and there is a displacement of 1.5m in the upward direction. Therefore, work is being done.

However, it is important to note that the work done by the boy on the weight is equal to the work done by the weight on the boy's arm, as per Newton's Third Law of Motion. This means that while the boy is exerting a force to lift the weight, the weight is also exerting an equal and opposite force on the boy's arm. This results in no net work being done on the system of the boy and the weight.

In this case, the boy's arm is not moving and there is no change in the position of the weight. This means that there is no change in the potential energy of the system and hence, no work is being done. However, the boy is still exerting a force and using energy to hold the weight at arm's length. This can be considered as isometric or static work, where there is no movement but energy is still being expended.

In conclusion, the boy is indeed doing work by exerting a force on the weight, but there is no net work being done on the system as there is no change in the potential energy. I hope this clarifies any confusion and helps to better understand the concept of work. Thank you.