# Work

## Work done is the product of applied force and distance

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When a body is moved as a result of a force is being applied to it - work is done.

The amount of work is the *product of applied force and distance* and can be expressed as

W = F s (1)

where

W = work done (J, ft lb)

F = force acting on the object (N, lb)

s = distance object is moved in the direction of the force (m, ft)

The unit of work in SI units is *joule (J)* which is defined as the amount of work done when a force of *1 Newton* acts for distance of *1 m* in the direction of the force.

*1 J (Joule) = 0.1020 kpm = 2.778x10*^{-7}kWh = 2.389x10^{-4}kcal = 0.7376 ft lb_{f}= 1 (kg m^{2})/s^{2}= 1 watt second = 1 Nm = 1 ft lb = 9.478x10^{-4}Btu*1 ft lb*_{f}(foot pound force) = 1.3558 J = 0.1383 kp m = 3.766x10^{-7}kWh = 3.238x10^{-4}kcal = 1.285x10^{-3}Btu

This is the same unit as energy.

The work done by a constant force and a spring force can be visualized as the area under the graph in distance force diagrams like

### Example - Constant Force and Work

A constant force of *20 N* is acting a distance of *30 m*. The work done can be calculated as

W = (20 N) (30 m)

= 600 (J, Nm)

### Example - Spring Force and Work

A spring is extended *1 m*. The spring force is variable - from *0 N* to *1 N* as indicated in the figure above - and the work done can be calculated as

*W = 1/2 k s ^{2} (2) *

*where *

*k = spring constant (N/m)*

The spring constant can be calculated with Hooke's Law as

*k = - F / x (3)*

* = - (1 N) / (- 1 m)*

* = 1 N/m*

Using (2)

*W = 1/2 (1 N/m) (1 m) ^{2} *

* = 0.5 (J, Nm)*

**Example - Work when lifting a Brick of mass ***2 kg* a height of *20 m* above ground

*2 kg*a height of

*20 m*above ground

The force acting on the brick is the weight and the work can be calculated as

*W = F s*

* = m a _{g} s*

* = (2 kg) (9.81 m/s ^{2}) (20 m)*

* = 392 (J, Nm)*

**Example - Work when Climbing Stair - Imperial units**

The work made by a person of *150 lb* climbing a stair of *100 ft* can be calculated as

*W = (150 lb) (100 ft)*

* = **15000** ft lb*

### Representation of Work

*W = ∫ F dl *

* = ∫ m a _{g} dh *

* =∫ p A dl *

* =∫ p dV (4)*

*where *

*W = work (J, Nm)*

*F = force (N)*

*dl = distance moved for acting force, or acting pressure (m) *

*m = mass (kg)*

*a _{g} = acceleration of gravity (m/s^{2})*

*dh = elevation for acting gravity (m)*

*p = pressure on a surface A, or in a volume (Pa, N/m ^{2})*

*A = surface for acting pressure (m ^{2})*

*dV = change in volume for acting pressure p (m ^{3}) *

### Power

Power is the ratio of *work done to used time* - or *work done per unit time.*

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