Work & Power
Overview
What is Work?

If an applied force causes a body to be
displaced in the direction of the force,
then work is done on the body by the
force.
• The work is done only while the force is being
applied, and
• The maximum work is done when the direction of
the force and the displacement are parallel.

Note! – no work if no displacement!
Work Formula and Units
• Work = applied force (F) x displacement (d) in
direction of force, or
• W = F x d x cos θ
• Where θ = the angle between the applied force and
• Units
the displacement direction
• Newton-meters (N-m), or
• Joules
• 1 N-m = 1 J
• Work is a scalar quantity
Two Analytical Models

Horizontal

Vertical
• The applied force causes horizontal motion
• The applied force is the gravitational force (or
weight) being overcome when an object is
being lifted vertically
Horizontal Motion Model

The force is applied
parallel (θ = 0º) to the
direction of motion
• W = F x d (if θ=0 then cos
θ = 1)

The force is applied at
an angle (θ) to the
direction of motion
• W = F x d x cos θ
Vertical (Lifting) Model
• The force is the gravitational force (weight)
being overcome by lifting in a vertical plane
• W = F x d, or
• W = mg x d, where
• m = mass of the body being lifted
• d = distance body is lifted
What is power?

Definition

Formula

Units
• The rate at which work is done.
• P = Work done/time taken
• P = W/t
• Watts (= joule/sec)
Power in Today’s Terms

USA uses British units – horsepower

Examples
• 1 hp = 746 watts
• 1 hp = 0.75 Kilowatt
• 100 hp Civic = 75 Kw
• 250 hp Subaru = 188 Kw
• 4000 hp Locomotive = 2900 Kw
Example 1

How much work is done when a 15 kg
crate is pushed across a floor with a
force of 100 N, displacing it 5 m.
Given

Find

W = F x d = 100 x 5 = 500 N-m (J)

• F = 100N, d = 5 m, m = 15 kg
• Work done (W)
Example 2

How much work is done when a 15 kg
crate is lifted 5 m?
Given

Find

W = F x d = mg x d
W = 15 x 9.8 x 5 = 735 N-m (J)


• d = 5 m, g = 9.8m/s/s, m = 15 kg
• Work done (W) (lifting model)
Example 3

How much work is done when a 15 kg
crate is pulled 5 m across a floor with a
force of 100N @ 30º to the horizontal?
Given

Find

• d = 5m, F = 100N, θ = 30˚, m = 15kg
• Work done (horizontal model)
• W = F x d x cosθ
• W = 100 x 5 x cos 30 = 433 N-m (J)
Example 4

How much power was expended in the
previous example if the pull lasted 5
seconds?
Given

Solve

• W = 433 J, t = 5 s, P = ?
• P = W/t = 433/5
• P = 86.6 W(atts)
Your turn!