Calculating Force in Lifting a 2.0kg Object

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    Force Lifting
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Homework Help Overview

The discussion revolves around calculating the work done in lifting a 2.0kg object from the bottom of a well at a constant speed. The subject area includes concepts of force, work, and motion in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the work done using the formula w=fd and seeks clarification on the force involved, questioning the acceleration of the system. Some participants question the implications of lifting at constant speed and the net forces acting on the object.

Discussion Status

The discussion is active, with participants exploring the relationship between constant speed and acceleration. Guidance has been provided regarding the implications of constant speed on net force and acceleration, but no consensus has been reached on the final calculations.

Contextual Notes

Participants are considering the effects of gravitational force and the applied force while discussing the assumptions related to the motion of the object.

je55ica7
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How much work is done by a person lifting a 2.0kg object from the bottom of a well at a constant speed of 2.0 m/s for 5.0s?

I know w=fd. I found the distance to be 10m. But what is the force? F=ma so what is the acceleration of the system?
 
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Since the object is being lifted at constant speed, what does that tell you about the acceleration? What's the net force on the object? (There are two forces acting on the object: the force applied by the person and the weight of the object.)
 
Wouldn't that mean that the acceleration is 0? So would I use w=mgh?
 
Yes and yes.
 

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