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Work
When a force moves something, work is done. Whenever work is done, energy is changed into a
different form.
Chemical energy Chemical energy → → Kinetic energyKinetic energy
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Work done by a constant force
Work = force in the direction x displacementof displacement
W = Fs
s
FF
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Where the force and displacement are not in the same direction, the component of force in the direction of force in the direction of displacementdisplacement is used.
s
FF
F cos F cos
F sin
W = (F cos ) s = Fs cos
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Positive, zero and negative work done 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.
Work done by the force = 5 x 2 = 10 J. Work done = + ve mechanical energy of the block is ⇒ increa
sed due to the force
2 m
5 N5 N
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A 5-kg suitcase is carried by a man on his shoulder for 3 m.
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
50 N 50 N
3 m
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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.
Work done by the force = -7500 x 40 = 300 000 JWork done = - ve- ve
⇒ mechanical energy of the block is decreaseddecreased due to the force
7500 N 7500 N
40 m
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Work done by a varying force
W is the work done by F during a very small displacement x. ⇒ W = Fx
The total work done = sum of all work done during all small displacements
= ∑Fixi = ∫Fdx= Area under force – displacement graph.
Force
Displacement
F
xs0
W
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EnergyEnergy exists in many different forms.
But we shall only study the different forms of mechanical energy.
Energy
Elastic potential energy
Gravitational potential energy
kinetic 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.
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.
Fat rest v
F
s
mm
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Kinetic Energy In general, if the velocity of a body of mass m increases
from from uu to to vv when work is done on it by a constant force F acting over a distance s,
Since the initial velocity = initial velocity = uu, by equation of motion, we have 2as = v2 – u2. Therefore, a = (vv2 2 – u– u22)/2s.
Kinetic energy gained by the body = Work doneWork done by F = Fs = (ma)s = m[(vv2 2 – u– u22)/2s]s = ½ mv2 –– ½ mu½ mu22.
F
uu v
F
s
mm
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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.
h
F
F
mg
mgmg
If the potential energy at the ground surface is taken to be zero, potential energy at the height h above the ground = work donework 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. FF ∝ xx or FF = kxx where k is the force constant
ExtensionExtension xx
Applied forceApplied force FF
Natural length l
If k = 100 N m-1, find the tension if the extension is 5 cm.T = k x = (100)(0.05)= 5 N
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Elastic potential energy
Elastic potential energy in the string
= work donework done by the force to achieve an extension x
= Area under the F – t graph
= ½ Fx = ½ (kx)x = ½ kx2
ExtensionExtension xx
Applied forceApplied force FF
Natural length l
Force
Extension
F
xF F ∝∝ xx
F = kx
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Conservation of energy
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
Mechanical energy is conserved
P.E. lost = K.E. gained
rough
Mechanical energy not conserved
P.E. lost > K.E. gained
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A block of mass 5 kg slides down an incline plane from A block of mass 5 kg slides down an incline plane from rest. If the angle of inclination is 30rest. If the angle of inclination is 30oo and the coefficient and the coefficient of kinetic friction of kinetic friction between the block and the plane is between the block and the plane is 0.2.0.2.
(a)(a) Determine work done by the gravitational force if Determine work done by the gravitational force if the distance traveled is by the block is 3 m.the distance traveled is by the block is 3 m.
(b)(b) Determine the corresponding work done by the Determine the corresponding work done by the friction.friction.
(c)(c) Hence, find the speed attained by the block.Hence, find the speed attained by the block.
30o
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Solution:Solution:
(a)work done by the gravitational force
= mg sin 30o x 3 = 75 J (P.E. loss by the block)
(b) frictional force = R = 0.2 x mg cos 30o = 8.66 N
work done by friction = -8.66 x 3 = -26.0 J
(Note: work done byby friction = -26.0 J; work done againstagainst friction = 26.0 J)
(c) K.E. gained = P.E. loss – work done against friction
= 75 – 26.0 = 49 J
½ mv2 – ½ mu½ mu22 = 49
½ (5)v2 – 0 = 49
v = 4.43 ms-1
mg cos 30omg sin 30o
mg
R
f
30o
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One end of an elastic string is connected to a fixed One end of an elastic string is connected to a fixed point A and the other end is connected to an object of point A and the other end is connected to an object of mass 2 kg as shown. If the object is released from A, mass 2 kg as shown. If the object is released from A, find the extension of the string when the object is find the extension of the string when the object is instantaneously at rest. It is given that the natural length instantaneously at rest. It is given that the natural length of the string is 30 cm and the force constant is 100 Nmof the string is 30 cm and the force constant is 100 Nm-1-1..
AA
t = 0
Natural length = 30 cm
Extension x ?
Instantaneously at rest
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Solution:By conservation of energy,P.E. lost = Elastic P.E. gained2(10)(0.3 + x) = ½(100)x2
50x2 – 20x – 6 = 0x = 0.6 m or x = -0.2 m (rejected)
AAAA
t = 0
Natural length = 30 cm
Extension x ?
Instantaneously at rest
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Conservative force 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 independ
ent of the path taken. i.e. WPath 1= WPath 2
a
bPath 1
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.
Work done by gravitational force = mg sin s = mgh
(independent of the angle of the slide)
hh
s
mgmg
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Conservative force Non-conservative force
Gravitational force Frictional force
Electrostatic force Air resistance
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Power The rate at which work is done or energy is transferred.
s
FF
Average power = work done / time taken
P = (Fs)/t = F(s / t) = Fv If the force acts on the body at an angle in direction of t
he motion,
s
FF
F cos
F sin
P = (Fs cos )/t = Fcos (s / t) = F cos v
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Each time the heart pumps, it accelerates about 20 g Each time the heart pumps, it accelerates about 20 g of blood from 0.2 msof blood from 0.2 ms-1-1 to 0.34 ms to 0.34 ms-1-1..(a)(a) What is the increase in kinetic energy of the What is the increase in kinetic energy of the blood with each beat? blood with each beat? (b)(b) Calculate the power of the heart when it beats Calculate the power of the heart when it beats at about 70 times per minute.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
%100InputPower
OutputPower UsefulEfficiency
%100inputEnergy
OutputEnergy UsefulEfficiency
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