Type: Single
Date: ______________
Objective: Work done by a constant force
Homework: Assignment (1-21)
Read (6.1, 6.2 and 6.7), Do CONCEPT QUEST
Ch. 6 #’s (2, 4) Ch. 6
AP Physics “B”
Mr. Mirro
W
Date:
ork Done by a Constant Force
________
Work is a familiar concept !
For example, it takes WORK to push a stalled car. In fact, the greater the “force” applied or the greater
the “position” change – then, the greater the mechanical “work” done.
Therefore, FORCE and DISPLACEMENT are two essential components of doing mechanical WORK.
The diagram shows a constant pushing force (F) that points in the same direction as the
resulting displacement (s).
The work (W) is defined as the magnitude (F) of the force times the magnitude (s) of the displacement.
That is:
W=Fs
Where (s) represents the displacement
such that horizontal displacement: sx ⇒ x
and vertical displacement: sy ⇒ y
The work done to push a car is the same whether the car is moved north to south or east to west,
provided that the amount of force used and the distance moved are the same. Since work does not
convey directional information, it is considered a scalar quantity.
The equation (W)ork = (F)orce x (D)istance indicates that the unit of work is derived from units of
force (N) and units of distance (m), or the Newton-meter.
One Newton-meter is equal to one Joule
1Nm = 1J
This implies that one joule (1 J) of work is done when a constant force of one newton (1 N) moves
an object through a distance of one meter (1 m).
The definition of work as W = F s does have one surprising feature:
If the distance (s) = 0 then the work (w) = 0. Even if a force (F) is applied !
Pushing on an immovable object such as a brick wall, may tire your muscles, but there is NO
MECHANICAL WORK done on the wall.
In physics, the idea of work is intimately tied up with the idea of motion !
Often, the force and displacement DO NOT point in exactly the same direction. In the diagram, the
force is directed at an angle ( θ ) relative to the displacement ( s ).
However, the component of the force that does the mechanical work is the horizontal
component F cos θ.
Now we can represent the general case for the work done by a constant force:
DefN: The work done (W) on an object by a constant force (F) is:
W = (F cos θ) s
Where F is the magnitude of the force, s is the magnitude of the
displacement, and θ is the angle between the force and the displacement.
AP Physics “B”
Mr. Mirro
W
Date:
ork Done by a Constant Force
________
Ex 1: A person pulls a toboggan for a distance of 25 m along the snow with a rope directed 20° above the
horizontal. If the tension in the rope is 85 N: [CutnellP6.3sim]
a. How much work is done on the toboggan by the tension force ?
b. How much work is done if the same tension is directed parallel to the snow ?
Ex 2: The drawing shows a boat being pulled by two locomotives through a canal of length 2.0 km. The
tension in each cable is 5.0 x 103 N, and θ = 20.0°. What is the net work done on the boat by the
two locomotives ? [CutnellP6.4]
Ex 3: The weight lifter in the figure is bench pressing a barbell whose weight is 710 N. In the first part
of the figure, he raises the barbell a distance of 0.65 m above his chest and in the second part he
lowers it the same distance. If the weight is raised and lowered at a constant velocity, determine
the work done on the barbell by the weight lifter during: [Cutnell 6.2]
a. the lifting phase
b. the lowering phase
Ex 4: A 2400 N force is pulling an 85 kg refrigerator across a horizontal surface. The force acts at an
angle of 20° above the surface. The coefficient of kinetic friction is 0.20, and the refrigerator
moves a distance of 8.0 m. Find: [Cutnell6.8]
a. The work done by the pulling force.
b. The work done by the kinetic frictional force.