# What is the work done in this case?

• vcsharp2003
In summary, the work done in stretching the spring by 15 cm is 5.627 J and it is independent of the magnitude of force applied. This is because work is determined by the change in kinetic energy and the sum of potential energies, both of which are independent of force applied. The work can also be calculated using the equation W = 1/2kx^2, where k is the spring constant and x is the displacement.
vcsharp2003

## Homework Statement

A spring (k = 500 N/m) supports a 400 g mass in a vertical position. The spring is now stretched 15 cm and maintained at the stretched position. How much work is done in stretching the spring by 15 cm? Is this work dependent on the magnitude of force applied while stretching the spring?

I am not sure if the work done would be just 0.5kx2 = 0.5 x 500 x 0.15[SUP}2[/SUP] = 5.625 J or one should use a more rigorous approach outlined under

.[/B]

## Homework Equations

For a spring,
F= kx
PEspring = 0.5kx2

Work energy Theorem
ΔKE + Σ ΔPE = Work done

## The Attempt at a Solution

Initial stretch of spring by 400 g mass = (0.400 x 9.8) / 500 = 0.00784 m
The initial stretch is used to determine initial spring PE.

Since the initial and final positions are at rest, so ΔKE = 0.
Σ ΔPE = ΔPEspring + ΔPEgravity = (0.5 x 500 x (0.15 + 0.00784)2 - 0.5 x 500 x (0.00784)2) + (0 - 0.400 x 9.8 x 0.15 )
= (6.23 - 0.015) + (- 0.588)
= 5.627 J
∴ Work done = ΔKE + Σ ΔPE = 0 + 5.627 = 5.627 J

This work would be independent of the magnitude of force applied since work done depends on only ΔKE and Σ ΔPE, both of which are independent of force applied.

With only one significant figure in your original measurements, I don't think the difference in your answers will matter.

So, the work in this context is force applied dotted with the displacement. You already recognize the force of the spring is ##F=kx##. We know that the displacement is 15 cm. Since the force changes as we stretch it further and further, we need to take the integral of the force with respect to x. When you do this, you'll end up with ##W=\frac 1 2 kx^2##.

## What is the work done in this case?

The concept of work in science refers to the transfer of energy from one system to another. It is calculated by multiplying the force applied to an object by the distance it moves in the direction of the force. In simpler terms, work is done when an object is moved by a force.

## How is work different from power?

While work is the transfer of energy, power is the rate at which work is done. It is calculated by dividing the work done by the time it takes to do it. In other words, power measures how quickly work is being done.

## Can work be negative?

Yes, work can be negative. This occurs when the force applied to an object is in the opposite direction of its movement. In this case, the work done on the object is considered to be negative.

## What are some real-life examples of work being done?

There are many examples of work being done in everyday life, such as lifting a book, pushing a shopping cart, or climbing stairs. In each of these cases, a force is applied to an object and it moves in the direction of the force, resulting in work being done.

## How is work related to kinetic and potential energy?

Work is directly related to kinetic and potential energy. When work is done on an object, it gains energy in the form of either kinetic energy (energy of motion) or potential energy (stored energy). Similarly, when work is done by an object, it loses energy in the form of kinetic or potential energy.

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