Work and energy

Objectives

1. Recognize the difference between the scientific and the ordinary definitions of work.

2. Define work, relating it to force and displacement.3. Identify where work is being performed in a

variety of situations.4. Calculate the net work done when many forces are

applied to an object.

Homework:

• In physics, work is defined as a force acting upon an object to cause a displacement.

Definition and Mathematics of Work

Work is being done

Work is not being doneWork is not being done

Let’s practice – work or no work1. A student applies a force to a wall and

becomes exhausted. 2. A calculator falls off a table and free falls to

the ground.3. A waiter carries a tray full of beverages

above his head by one arm across the room

4. A rocket accelerates through space.

no work

work

no work

work

Calculating the Amount of Work Done by Forces

θF

d

• F - is the force in Newton, which causes the displacement of the object.

• d - is the displacement in meters• θ = angle between force and

displacement• W - is work in N m or Joule (J). 1 J = 1 ∙

N m = 1 kg m∙ ∙ 2/s2

• Work is a scalar quantity • Work is independent of time the force

acts on the object.θF

d

Fx

Fy

Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.

cosFdW

Example 1• How much work is done on a vacuum cleaner

pulled 3.0 m by a force of 50.0 N at an angle of 30o above the horizontal?.

cosFdW omNW 30cos)0.3)(0.50(

JW 130

Example 2• How much work is done in lifting a 5.0 kg box

from the floor to a height of 1.2 m above the floor?

W = F dcos∙ θF = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 NW = F d ∙ = (49 N) (1.2 m) = 59 J

Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0Unknown: W = ?

Example 3• A 2.3 kg block rests on a horizontal surface. A constant force of

5.0 N is applied to the block at an angle of 30.o to the horizontal; determine the work done on the block a distance of 2.0 meters along the surface.

Given: F = 5.0 N; m = 2.3 kg

d = 2.0 m θ= 30o

30o

5.0 N

2.3 kgunknown:

W = ? J Solve:W = F d cos∙ ∙ θW = (5.0 N)(2.0 m)(cos30o) = 8.7 J

Example 4• Matt pulls block along a horizontal surface at constant

velocity. The diagram show the components of the force exerted on the block by Matt. Determine how much work is done against friction.

8.0 N

6.0 N

3.0 m

W = Fxdx

W = (8.0 N)(3.0 m) = 24 J

Given: Fx = 8.0 NFy = 6.0 N dx = 3.0 m

unknown: W = ? J F

Class work

• Page 170 practice #1-4

1. 1.50 x 107 J2. 7.0 x 102 J3. 1.6 x 103 J4. 1.1 m

The sign of work

When No work is done

0cos FdW

Force vs. displacement graph• The area under a force versus displacement

graph is the work done by the force.

Displacement (m)Fo

rce

(N)

work

Example: a block is pulled along a table with 10. N over a distance of 1.0 m.

W = Fd = (10. N)(1.0 m) = 10. J

height base area

The angle in work equation

• The angle in the equation is the angle between the force and the displacement vectors.

F & d are in the same direction, θ is 0o.

Fd

What is θ in each case?

Class work

• Page 171 #1-6