Work, Energy and Power

When a body which a force is acting moves in the direction of the force, work is done and energy
transformed. Work is defined as the product of force and distance, and though force is a vector,
work itself is a scalar quantity. The unit of work and energy is the joule (J). Consider a body moving
in a horizontal direction of motion exerted on a body by some outside agent.

Fig. 1. The work done, w = P cos × s; s = displacement, P cos = component of the force, F along
the horizontal direction.

When several external forces act on a body, the total work is the algebraic sum of the individual
works. As an illustration consider a box being dragged along a horizontal surface by a constant
force, P making a constant angle in the direction of motion.

The other forces acting on the box are (i) its weight (ii) normal upward force, N exerted by the
force and (iii) the frictional force, f. The work done by each force when the box moves a distance,
s along the surface to the right can be calculated as follows;

Fig 2 an object on a rough horizontal surface moving to the right under the action of force P,
incline at an angle . The component of p in the direction of the motion is P cos . The work of
the force, P is WP = (P cos ) × s …………………………… (2)

The forces w and N are both at right angle (90°) to the displacement. Hence, Ww =0}(because cos
90=0) ………………..(3)

The frictional force, f is the opposite of the displacement so the work of friction force is

Wf = -f s ……………………………… (4)

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Energy sources

Energy is the ability to do work; it is a scalar quantity with the same unit as work (J). Energy is
available from a number of sources:

1. Fossil fuels: from coal, oil, gas.

2. Other forms; wind, waves, electricity (hydro), wood, nuclear fission and fusion
geothermal (earthquake), solar energy, biochemical energy, tides.
Energy exists in the following forms
a. Mechanical (K.E or P.E).
b. Chemical
c. Electrical
d. Magnetic
e. Sound

Kinetic Energy
Consider a constant resultant force F. Such a force, acting on a particle of mass m, will produce a
constant acceleration a. let us choose the x-axis to be in the common direction of F and a. In order
to determine the work done by this force on the particle, consider the relation.

One-half the product of the mass of a body and the square of its speed is the kinetic energy
represented by K. E = 1/2mv2

Potential Energy: it is the energy associated with the position of a body; in a gravitational field is
the gravitational potential energy of the body P.E =mgh where m = mass, g= acceleration due to
gravity; h= height.

POWER

It is the rate at which work is done. Like work and energy, power, is a scalar quantity. When a
quantity of work ∆𝑊 is done during a time interval ∆𝑡, the average work done per unit time or
average power Pav is Pav = ∆𝑊/∆𝑡

Instantaneous power p is the limit of average as ∆𝑡 approaches zero.

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The SI unit of power is the watt (W).

1 w = 1 𝐽⁄𝑠

1 KW = 103 W

1 MW = 106W

A larger unit of power called the horse power (hp) is also used, 1 hp = 746 W = 3/4 of a kilowatt.
In mechanics we can also express power in terms of force and velocity. Suppose that a force F
acts on a body while it undergoes a vector tangent to the path (parallel to ∆𝑠). Then, the work
done by the force is ∆𝑊 = F11∆𝑠, and the average power is

Instantaneous power P is the limit of this expression as ∆𝑡 0 P = F11v. Where v is the magnitude
of the instantaneous velocity. We can also express the above equation in terms of the scalar

product P = F · v

Examples:

1. A man uses a horizontal force of 200 N to push a crate that is 20o above the horizontal
(a) how much work does the man perform? (b) if the man takes 12s to push the crate up the
ramp, what is his power output in watts?

2. A car develops 50KW and moves with a uniform velocity of 30m/s. calculate (i) the
energy expended in 5sec. (ii) The frictional force it overcomes.

MOMENTUM AND IMPULSE

Consider a particle of mass m moving with acceleration a, Newton’s second law can be written
as

We can take the mass m inside the derivative because it is a constant.

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Thus Newton’s second law says that the net force ∑ F acting on the particle equals the time rate
of change of the combination of mv; the product of the particle’s mass and velocity. We will call
this combination the momentum, or linear momentum, of the particle using the symbol P
for momentum P = mv

Momentum is a vector quantity. Its unit is kgm/s. Consider a particle acted upon by a constant
net force ∑ 𝐹 during a time interval ∆𝑡 𝑓𝑟𝑜𝑚 t1 to t2. The impulse of the net force, denoted by J
is defined as the product of the net force and the time interval.

J = ∑ 𝐹 (t2 – t1) =∑ F ∆t

The second law of motion can be restated in terms of momentum, as

Comparing with the above equation for J in 3, we end up with a result called the impulse–
momentum theorem:

The change in momentum of a particle during a time interval equals the impulse of the net
force that acts on the particle during that interval.

Example 3: what is the magnitude of the momentum of a 10,000 kg truck whose speed is 12.0
m/s? (b) What speed should a 2000 kg sport utility vehicle have to attain in order to have (i) the
same momentum (ii) the same kinetic energy as in a above?

Principle of Conservation of Linear Momentum

This states that, the total momentum of an isolated system is always conserved

Features of collision

a. It occurs in a short time interval

b. What happens after the collision differs from what happens before collision
c. The colliding bodies may be assumed to constitute a closed system
d. Momentum and energy are conserved during the collision
Direct central impact

We consider two masses m1 and m2, initially moving with velocities v1 and v2 respectively. If after
impact, they have velocities v1' and v 2' respectively, then we state the law as follows:

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m1v1  m2 v2  m1v1'  m2 v 2' = constant

If, however, the bodies sticked together after impact and continue with a common velocity v 0,
then the law becomes:

m1v1  m2 v2  (m1  m2 )v0

A collision in which the total kinetic energy after the collision is less than before the collision is
called an inelastic collision. A meatball landing on a plate of spaghetti and a bullet embedding
itself in a block of wood are examples of inelastic collisions. An inelastic collision in which the
colliding bodies stick together and move as one body after the collision is often called a
completely inelastic collision.

1 1 1
m1v1  m2v2  (m1  m2 )v0
2 2 2
The kinetic energy for a completely inelastic collision is
2 2 2

If the second body was at rest before the collision then v2  0 and m1v1  (m1  m2 )v0 and
m1v1
v0 
m1  m2

In this case the kinetic energy k1 and k2 before and after collision becomes:

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K1 
2
m1v1 and
2

1 1 m1
K2  (m1  m2 )v0  (m1  m2 )(
2 2
) 2 v1
2 2 m1  m2

K2 m1
And ( )
K1 m1  m2

The right hand side of the ratio of K1 and K2 is always less than unity which indicates energy loss.

Example: An object of mass 8 kg moving with a velocity of 50 m/s collides with an object of mass
2 kg moving in the opposite direction with a velocity of 30m/s. after impact, the two bodies unite
and move with a common velocity, v. calculate the (a) value of v (b) energy loss.

ELASTIC COLLISION

If the forces between the bodies are also conservative, so that no mechanical energy is lost or
gained in the collision, the total kinetic energy of the system is the same after the collision as
before. Such a collision is called an elastic collision. In elastic collision both momentum and K.E is
conserved. For example, when two billiard balls collide, they spring back from every form of
deformation they undergo. From conservation of momentum,

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m1v1  m2 v2  m1v1'  m2 v 2' and the conservation of K.E gives

1 1 1 2 1 2
m1v1  m2 v1  m1v1'  m2 v2'
2 2
(2)
2 2 2 2

If body 2 was initially at rest before the collision, then the kinetic energy and momentum
conservation equations are:

1 1 2 1 2
m1v1  m1v1'  m2 v2'
2
(2)
2 2 2

And m1v1  m1v1'  m2 v2' (3

2 2 2 2
Solving equation 2 and 3 we get m2 v2'  m1 (v1  v1' )  m1 (v1  v1' )(v1  v1' )
2 2 2
(4)

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m2 v1'  m1 (v1  v1' ) (5)

Divide equation 4 by 5 to get v1'  (v1  v1' ) (6)

Substitute equation 6 into 5 to eliminate v1'

m1  m1
v1  ( ) (7)
m1  m2

Putting 7 into 6

2m1
v1'  ( )v1 (8)
m1  m2

Example 5: the figure below shows an elastic collision of two pucks on a frictionless air-hockey
table. Puck A has mass mA = 0.5 kg and puck B has mass mB = 0.3 kg. Puck A has an initial velocity
of 4.00 m/s in the positive direction and a final velocity of 2.00 m/s in an unknown direction. Puck
B is initially at rest. Find the final speed of puck B and the angles 𝛼 𝑎𝑛𝑑 𝛽 in the figure.

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