King Saud University College of Science Physics ...fac.ksu.edu.sa/sites/default/files/ch_7.1.pdfنارضخلا فون .أ 2 The work W done on a system by an agent exerting a constant

PHYS 103 (GENERAL PHYSICS)

CHAPTER 7:Energy and Energy Transfer

King Saud University

College of Science

Physics & Astronomy Dept.

Presented by Nouf Saad Alkathran

نوف الخضران. أ 2

The work W done on a system by an agent

exerting a constant force on the system is

the product of the magnitude F of the force,

the magnitude ∆r of the displacement of the

point of application of the force,

and cos Ɵ where Ɵ is the angle between

the force and displacement vectors:


Which force does work on this block?

The sign of the work also depends on the

direction of F relative to ∆r.

The work done by the applied force is

positive when the projection of F onto ∆r is in

the same direction as the displacement.

When the projection of F onto ∆r is in the

direction opposite the displacement, W is

negative.

The factor cos Ɵ in the definition of W automatically takes

care of the sign.

SI unit of work is the newton.meter (N·m) = the joule ( J).


Work is an energy transfer. If W is the work done on a

system and W is positive, energy is transferred to the system;

if W is negative, energy is transferred from the system.


The scalar product of any two vectors A and B is a scalar

quantity equal to the product of the magnitudes of the

two vectors and the cosine of the angle between them:



Consider a particle being displaced along the x axis under the

action of a force that varies with position. we cannot use W =

F ∆r cosƟ

Imagine that the particle undergoes a very small dis -

placement ∆x, the x component Fx of the force is

approximately constant over this small interval



This is just the area of the shaded rectangle. If we imagine

that the Fx versus x curve is divided into a large number of

such intervals, the total work done

If more than one force acts on a system and

the system can be modeled as a particle, the

total work done on the system is just the work

done by the net force


The work done by the force is equal to the

area under the curve from A=0 to C =6 m.

This area is equal to the area of the

rectangular section from A to B plus the

area of the triangular section from B to C.

The area of the rectangle is (5.0 N)(4.0 m) = 20 J, and

the area of the triangle is ½ (5.0 N)(2.0 m) = 5.0

J. Therefore, the total work done by the force on the

particle is 25 J.


For many springs, if the spring is either stretched or compressed a

small distance from its unstretched (equilibrium) configuration, it

exerts on the block a force


Suppose the block has been pushed to the left to a position 2x max

and is then released. We identify the block as our system and calculate

the work Ws done by the spring force on the block as the block moves

from xi =-x max to xf =0


let us consider the work done on the block by an external agent as

the agent applies a force on the block, the applied force 𝐹𝑎𝑝𝑝 is

equal in magnitude and opposite in direction to the spring force 𝐹𝑠


The work done by this applied force

(the external agent) on the system of

the block is


A common technique used to measure the force

constant of a spring is demonstrated by the

setup in Figure. The spring is hung vertically,

and an object of mass m is attached to its

lower end. Under the action of the “load” mg,

the spring stretches a distance d from its equilibrium

position. (A) If a spring is stretched 2.0 cm by a

suspended object having a mass of 0.55 kg,

what is the force constant of the spring?


(B) How much work is done by the spring as it stretches

through this distance?



Work–kinetic energy theorem:

When work is done on a system and the only change in the system

is in its speed, the net work done on the system equals the change

in kinetic energy of the system,

The work–kinetic energy theorem indicates that the speed of a

system increases if the net work done on it is positive because the

final kinetic energy is greater than the initial kinetic energy. The

speed decreases if the net work is negative because the final

kinetic energy is less than the initial kinetic energy.


Suppose the magnitude of the force in this example is doubled to

𝐹− =2F. The 6kg block accelerates to 3.5 m/s. How does the

displacement ∆𝑥− compare with the original displacement ∆x?

King Saud University College of Science Physics ...fac.ksu.edu.sa/sites/default/files/ch_7.1.pdfنارضخلا فون .أ 2 The work W done on a system by an agent exerting a constant

Documents