# Work

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

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 lbf = 1 (kg m2)/s2 = 1 watt second = 1 Nm = 1 ft lb = 9.478x10-4 Btu
• 1 ft lbf (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 s2                                               (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

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

W = F s

= m ag s

= (2 kg) (9.81 m/s2) (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 ag 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)

ag = acceleration of gravity (m/s2)

dh = elevation for acting gravity (m)

p = pressure on a surface A, or in a volume (Pa, N/m2)

A = surface for acting pressure (m2)

dV = change in volume for acting pressure p (m3)

### Power

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

## Related Topics

• Dynamics - Motion - velocity and acceleration, forces and torques
• Mechanics - Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more
• Thermodynamics - Effects of work, heat and energy on systems

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