Transcript 3 ms -1 - GEOCITIES.ws

Understanding Work, Energy,
Power & Efficiency
Energy
Work done is equal to the amount of energy transferred.
Forms of energy:
1. Kinetic energy
2. Gravitational potential energy
3. Elastic potential energy
4. Sound energy
5. Heat energy
6. Light energy
7. Electrical energy
8. Chemical energy
Energy – Work Done
When energy changes from
one form to another, Work
is done.
Energy transfer when the work
is done.
Examples
A student pushes a trolley
of books. Chemical energy
stored in the body of the
student is transferred to
the trolley as kinetic
energy.
The farmer lifting a bag of
rice onto the lorry transfer
energy to the bag as
gravitational potential
energy.
Only moving objects possess kinetic energy.
The kinetic energy of the object is the
same as the work done to change the
velocity of the object from 0 m s-1 to v
m s-1.
Initial velocity, u = 0 m s-1
Final velocity, v = v m s-1
From equation of linear motion:
Displacement, s = s m
V2 = u2 + 2as
Work done, W = F x s = m a s
S = (v2) / 2a
W = ma (v2/2a)
W = ½ m v2
The kinetic energy, Ek of an
object of mass, m kg traveling
at a velocity of v m s-1 is given
as:
Ek = ½ m v2
EXAMPLE 1
A car of mass 950 kg accelerates from a velocity
of 20 m s-1 to a velocity of 35 m s-1. What is the
work done for the car to accelerate?
v = 35 ms-1
Solution:
m = 950 kg, u = 20 ms-1, v = 35 ms-1
Work done, W = Change in kinetic energy of the car
W = Final Ek – Initial Ek
W = ½ mv2 – ½ mu2 = ½ m (v2-u2)
W = ½ x 950 ms-1 x (35 ms-1 – 20 ms-1) = 391875 J
EXAMPLE 2
What is the kinetic energy of a man of mass
65 kg jogging at a velocity of 3 m s-1
Solution
Mass, m = 65 kg, velocity, v = 3 m s-1
Kinetic energy, Ek = ½ mv2
Ek = ½ x 65 kg x (3 m s-1)2
Ek = 292.5 J
When an object of mass, m kg is raised to a
height, h meters above the earth’s surface,
the object possesses gravitational potential.
box
h
The gravitational potential energy is equal to the
work done to raise an object to a particular height.
Work done, W = F x s
W = mgh
Gravitational potential energy, Ep = W
Ep = mgh
EXAMPLE 1
Figure below shows a box of weight 50 N being pulled
up on a smooth inclined plank by a force of 25 N
from A to B.
box
4m
B
50 N
h=2m
30o
A
a. Calculate the work done by the 25 N force.
b. What is the work done against gravity?
c. State the value of the gravitational potential energy of the box at B.
EXAMPLE 2
Calculate the gravitational potential energy in
respect of each of the following.
m = 2 kg
h = 1.2 m
Solution
Potential energy of weight, Ep = mgh
Ep = 2 kg x 9.8 ms-2 x 1.2 m
Ep = 23.52 J
The total amount of energy always remains the same.
EXAMPLE 1
In a softball game, a ball was miss hit and flew
vertically upwards with an initial velocity of 15 ms-1.
what is the maximum height attained by the ball?
Solution
If air resistance is neglected, we can assume that the kinetic
energy of the ball is changed to gravitational potential energy
only.
½ m v2 = mgh
½ x m x (15 ms-1) 2 = m x 9.8 ms-2 x h
h = (15 ms-1)2 / (2 x 9.8 ms-2)
h = 11.48 m
EXAMPLE 2
A durian of 2.5 kg falls from a height of 12 m to the ground.
a. What is the loss of potential energy when the durian is 4
meters above the ground?
b. Calculate the velocity of the durian at this height.
Solution
a. Loss of potential energy = m g (12 m) – m g (4 m)
= m g (12 m – 4 m) = 2.5 kg x 9.8 ms-2 x 8 m
= 196 J
b. Gain in kinetic energy = Loss of potential energy
= 196 J
½ x 2.5 kg x v2 = 196 J
v2 = (2 x 196 J) / (2.5 kg) = 156.8
v = 12.5
Quiz
Calvin rides his bicycle down the slope of a hill 3 m high at an
initial velocity of 2 m s-1, without pedaling. At the foot of the hill,
the velocity is 6 m s-1. given that the mass of Calvin with his
bicycle is 75 kg, find
a.The initial kinetic energy of the bicycle,
b.The initial potential energy of the bicycle,
c.The work done against friction along the slope.
u = 2 m s-1
v = 6 m s-1
3m
Solution
a. Initial kinetic energy, Ek = ½ m u2
Ek = ½ (75 kg) x (2 m s-2) = 150 J
b. Initial potential energy, Ep = mgh
Ep = (75 kg) x (9.8 m s-2) x (3 m)
Ep = 2205 J
c. Kinetic energy at the foot of the hill, Ek = ½ m v2
Ek = ½ (75 kg) x (6 m s-1)2
Ek = 1350 J
Work done against friction along the slope,
W = total initial energy – kinetic energy at the foot of the hill
W = (150 + 2205) – 1350
W = 2355 – 1350 = 1005 J
Summary