Transcript Projectile motion

Work
Chemical energy → Kinetic energy


When a force moves something, work is done.
Whenever work is done, energy is changed into a
different form.
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Work done by a constant force
F
s


Work = force in the direction x displacement
of displacement
W = Fs
2

Where the force and displacement are not in the same
direction, the component of force in the direction of
displacement is used.
F sin q
F
q
F
q
F cos q
s

W = (F cos q) s = Fs cos q
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Positive, zero and negative work
done
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A block is initially at rest and placed on a smooth floor. It
is pushed by a horizontal force of 5 N for 2 m.
5N
5N
2m


Work done by the force = 5 x 2 = 10 J.
Work done = + ve ⇒ mechanical energy of the block
is increased due to the force
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
A 5-kg suitcase is carried by a man on his shoulder for
3 m.
50 N
50 N
3m
Work done by the force = (50 cos 90o)(3) = 0 J
Work done = 0
⇒ mechanical energy of the block remains unchanged
due to the force
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A car is traveling and brakes are applied to stop the car.
The braking force is 7500 N and the braking distance
is 40 m.
7500 N
7500 N
40 m
Work done by the force = -7500 x 40 = 300 000 J
Work done = - ve
⇒ mechanical energy of the block is decreased due to
the force
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Work done by a varying force
Force
dW
F
0


dx
s
Displacement
dW is the work done by F during a very small displacement
dx. ⇒ dW = Fdx
The total work done = sum of all work done during all small
displacements
= ∑Fidxi = ∫Fdx
= Area under force – displacement graph.
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Energy
Energy exists in many different forms.
But we shall only study the different forms of mechanical
energy.
kinetic energy
Gravitational
potential energy
Elastic potential
energy
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Kinetic Energy

A moving object has kinetic energy. Consider a body of
mass m which is initially at rest. Let a constant force F
act on it over a distance s and bring it to move with
velocity v.
v
at rest
m
F
m
F
s
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
Since the initial velocity = 0, by equation of motion, we
have 2as = v2 – 02. Therefore, a = v2/2s.
Kinetic energy of the body = Work done by F
= Fs = (ma)s = m(v2/2s)s = ½ mv2.
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Kinetic Energy

In general, if the velocity of a body of mass m increases
from u to v when work is done on it by a constant force
F acting over a distance s,
u
v
F
F
m
m
s


Since the initial velocity = u, by equation of motion, we
have 2as = v2 – u2. Therefore, a = (v2 – u2)/2s.
Kinetic energy gained by the body = Work done by F
= Fs = (ma)s = m[(v2 – u2)/2s]s = ½ mv2 – ½ mu2.
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Gravitational potential energy


Gravitational potential energy is the energy an object
possesses because of its position above the ground.
Consider an object of mass m being lifted vertically for a
height h from the ground.
F
mg
F
mg
h
If the potential energy at the
ground surface is taken to be zero,
potential energy at the height h
above the ground
= work done by the force
= Fs = mgh
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Elastic potential energy

Hooke’s Law:
For an elastic string or spring, the extension x is
directly proportional to the applied force F if the elastic
limit is not exceeded. i.e. F ∝ x or F = kx where k is the
force constant
Natural
length l
Extension x
If k = 100 N m-1,
find the tension if the
extension is 5 cm.
T = k x = (100)(0.05)
=5N
Applied force
F
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Elastic potential energy
Force
Natural
length l
F = kx
F
Extension x
F ∝x

Applied force
F
x
Extension
Elastic potential energy in the string
= work done by the force to achieve an extension x
= Area under the F – t graph
= ½ Fx = ½ (kx)x = ½ kx2
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Conservation of energy
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
Energy may be transformed from one form to another,
but it cannot be created or destroyed, i.e. the total
energy of a system is constant.
The total amount of mechanical energy (K.E. + P.E.) is
constant unless the motion is frictionless.
i.e. K.E. lost = P.E. gained or P.E. lost = K.E. gained
smooth
rough
Mechanical energy is conserved
Mechanical energy not conserved
P.E. lost = K.E. gained
P.E. lost > K.E. gained
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A block of mass 5 kg slides down an incline plane from
rest. If the angle of inclination is 30o and the coefficient
of kinetic friction m between the block and the plane is
0.2.
30o
(a)
(b)
(c)
Determine work done by the gravitational force if
the distance traveled is by the block is 3 m.
Determine the corresponding work done by the
friction.
Hence, find the speed attained by the block.
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R
Solution:
f
mg sin 30o
30o
mg cos 30o
mg
(a) work done by the gravitational force
= mg sin 30o x 3 = 75 J (P.E. loss by the block)
(b) frictional force = mR = 0.2 x mg cos 30o = 8.66 N
work done by friction = -8.66 x 3 = -26.0 J
(Note: work done by friction = -26.0 J; work done against friction = 26.0 J)
(c) K.E. gained = P.E. loss – work done against friction
= 75 – 26.0 = 49 J
½ mv2 – ½ mu2 = 49
½ (5)v2 – 0 = 49
v = 4.43 ms-1
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One end of an elastic string is connected to a fixed point
A and the other end is connected to an object of mass 2
kg as shown. If the object is released from A, find the
extension of the string when the object is
instantaneously at rest. It is given that the natural length
of the string is 30 cm and the force constant is 100 Nm-1.
A
A
Natural length
= 30 cm
t=0
Extension x ?
Instantaneously
at rest
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A
A
Natural length
= 30 cm
t=0
Extension x ?

Instantaneously
at rest
Solution:
By conservation of energy,
P.E. lost = Elastic P.E. gained
2(10)(0.3 + x) = ½(100)x2
50x2 – 20x – 6 = 0
x = 0.6 m or x = -0.2 m (rejected)
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Conservative force
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Conservative force is a force whose work is determined
only by the final displacement of the object acted upon.
The total work done by a conservative force is
independent of the path taken.
i.e. WPath 1= WPath 2
Path 1
b
a
Path 2
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Conservative force

For example, if a child slides down a frictionless slide,
the work done by the gravitational force on the child from
the top of the slide to the bottom will be the same no
matter what the shape of the slide; it can be straight or it
can be a spiral. The amount of work done only depends
on the vertical displacement of the child.
s
h
h
mg
q
Work done by gravitational force = mg sin q s = mgh
(independent of the angle of the slide)
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Conservative force
Non-conservative force
Gravitational force
Frictional force
Electrostatic force
Air resistance
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Power
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The rate at which work is done or energy is transferred.
F
F
s
Average power = work done / time taken
P = (Fs)/t = F(s / t) = Fv
If the force acts on the body at an angle q in direction of
the motion,
F
F sin q F
q

q F cos q
s
P = (Fs cos q)/t = Fcos q (s / t) = F cos q v
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
Each time the heart pumps, it accelerates about 20 g
of blood from 0.2 ms-1 to 0.34 ms-1.
(a) What is the increase in kinetic energy of the
blood with each beat?
(b) Calculate the power of the heart when it beats at
about 70 times per minute.
Solution:
(a) Increase in K.E. = ½ mv2 – ½ mu2
= ½ (0.02)(0.34)2 – ½ (0.02)(0.2)2 = 7.56 x 10-4 J
(b) Power of the heart = energy / time
= (7.56 x 10-4 x 70) / 60
= 8.82 x 10-4 W
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Efficiency
Useful Power Output
Efficiency
100%
Power Input
Useful EnergyOutput
Efficiency
100%
Energyinput
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