Concepts
11 min read
Work, Energy And Power
- All important concepts of this chapter, in one place
1
Work done as dot product of force and displacement
Work is force applied over distance. For e.g., lifting an object against the Earth's gravitation, driving a car up a hill
etc.
W = Fd
The standard unit of work is the joule (J), which is equal to newton - meter (N m).
If force is not applied in the same direction as the displacement, then also it results in work (for example, a car
ascending a hill).
In this case, we will have to use the vector form of the above formula.
For a given force vector F (in newtons, in a given direction) applied to an object that undergoes displacement d (in
meters), the work w (in joules) is given by the dot product of the force vector and the displacement vector as :
W = F .d
W = F dcosθ
2
Work Done as area under force Vs displacement graph
Example: A force acts on a body and displaces it in it's direction. The graph
shows the relation between the force and displacement. The work done by the force is:
Work done by force =∫ F .ds =Area under F.s graph= Area of △
=
1
× base × height
2
=
1
× (14 − 2) × 60
2
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= 360 J
3
Work Energy Theorem
The work-energy theorem states that the work done on an object by the net force is equal to the change in its
kinetic energy.
Wnet = (K.E.)f − (K.E.)i
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4
Potential energy with respect to a reference line
Potential energy is always defined with respect to a reference line in space. Usually infinity is taken as the
reference line and potential energy defined is zero at infinity. In gravitation, for objects close to surface of earth,
ground surface is taken as reference line and energy at ground surface is zero.
Note:
Choice of reference line is not fixed and can be redefined. This helps in solving of problems.
5
Relationship between force and potential energy
Force is equal to the gradient of potential energy.
F =−
δU ^ δU ^ δU ^
i−
j−
k
δx
δy
δz
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Example:
A particle of mass 1 kg is moving in the x − y plane and its potential energy U in joule obeys the law U = 6x + 8y ,
where (x, y ) are the coordinates of the particle in meter. If the particle starts from rest at (9,3) at time t = 0, then
find the force on the body.
Solution:
F =−
δU ^ δU ^ δU ^
i−
j−
k
δx
δy
δz
​
F = −6^i − 8^j
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6
Work Done by ideal springs
Example: A spring obeying the linear law F = −Kx is first compressed by
10 cm and the work done is W1 . Next it is compressed by
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another 10 cm, the work done now is W2 , then what is the value of W1 : W2 ?
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Solution:
W1 = Δ(P .E)
​
1
K(x2 2 − x1 2 )
2
1
= K(102 − 02 )
2
=
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Similarly, W2 =
​
1
K(202 − 102 )
2
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So, W1 : W2 = 100 : 300
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=1:3
7
Conservative and Non-Conservative Forces
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If work done by any force in a closed path is zero then the force is conservative else it would be non-conservative.
For this defining any arbitrary closed path work done has to be calculated.
For the closed path:
∮ F .dx = 0
8
Conservation of total mechanical energy
Mechanical energy is the sum of the potential and kinetic energies in a system. The principle of the conservation
of mechanical energy states that the total mechanical energy in a system (i.e., the sum of the potential plus kinetic
energies) remains constant as long as the only forces acting are conservative forces.
Example: Consider a person on a sled sliding down a 100 m long hill on a 30 0 incline. The mass is 20 kg , and the
person has a velocity of 2 m/s down the hill when they're at the top. How fast is the person traveling at the bottom
of the hill?
Solution:
At the top: P .E = mgh = (20)(9.8)(100sin30^0) = 9800 J
KE = 1/2mv 2 = 1/2 × (20) × (2)2 = 40J
Total mechanical energy at the top = 9800 + 40 = 9840 J
At the bottom: P E = 0, K.E. = 1/2mv 2
Total mechanical energy at the bottom = 1/2mv 2
If we conserve mechanical energy, then the mechanical energy at the top must equal what we have at the bottom.
This gives:
1/2mv 2 = 9840, v = 31.3m/s.
9
Power
Power is defined as the rate of doing work. It also equals the amount of energy consumed in unit time. It is a scalar
quantity.
W
∴P =
t
Power can also be found as the product of force and velocity in the direction of force.
∴ P = Fv
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10
Problem based on conservation of momentum in elastic collision
Example: A ball of mass M moving with a velocity V collides head on elastically with another of same mass but
moving with a velocity v in the opposite direction. What will be the scenario after collision?
Solution:
Using conservation of momentum we have
M1 u1 − M2 u2 = −M1 v1 + M2 v2
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or
M V − M u2 = −M v + M v2
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or
V − u2 = −v + v2 .............(i)
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and as for a perfect elastic collision we have e=1, thus we get
v2 + v = u2 + V .................(ii)
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Using these two equations we get
u2 = v and v2 = V
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Thus the velocities are exchanged between the two balls.
11
Problem on perfectly inelastic collision in one dimension
Example: A particle of mass m moving in the direction x with speed 2v is hit by another particle of mass 2m
moving in the y direction with speed v . If the collision is perfectly inelastic, what is the percentage loss in the
energy during the collision?
Solution:
as the collision is inelastic the masses will move together
Assuming the speed of block A and B becomes v1^i + v2^j
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Writing momentum equation in x direction
2v
m(2v) + 2m(0) = 3mv1 ⇒ v1 =
3
Writing momentum equation in Y direction
2v
m(0) + 2m(v) = 3mv2 ⇒ v2 =
3
2v ^ 2v ^
The velocity of both blocks will be
i+
j
3
3
The loss in kinetic energy
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1
1
1
2 2v 2
m(2v)2 + 2m(v)2 − 3m(
)
2
2
2
3
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4
5
3mv 2 − mv 2 = mv 2
3
3
5 2
mv
percentage loss 3 2 × 100 = 56 %
3mv
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