Contents
Physics Topics can also be used to explain the behavior of complex systems, such as the stock market or the dynamics of traffic flow.
Is Mechanical Energy Always Stays Constant and Conserved?
Definition: A system in which the total mechanical energy i.e., the sum of its kinetic and potential energy remains conserved, is called a conservative system. The forces acting in such a system are called conservative forces.
Example: Gravitational force, restoring force of spring or elastic force, electrostatic force, force between the magnetic poles, etc., are conservative forces.
Work is done to lift a body against gravity and is stored as potential energy While returning to its initial position, the body uses this stored energy and does work which is exactly equal to the initial work done while lifting. Hence energy is conserved.
Work done in a closed path under a conservative force: For a conservative system, the work done depends only on the initial and final positions. It does not depend on the path taken in reaching the final position from the initial position. This means that the sum of the kinetic and potential energies remains constant throughout. It also implies that if a body goes around a complete loop so that its final and initial positions are the same, then the total work done is zero. For any conservative force, the work done is reversible. Hence if work done against a force can be restored, the force is called conservative. Also if work done by a force in a closed path equals to zero, the force is called conservative.
Thus in a conservative system, it is possible to restore the initial work done.
Suppose a body gets displaced by dx along the x-axis under the action of a conservative force F (which may vary with position) also acting along the x -axis. Therefore the work done by the conservative force dW = F(x)dx.
Since change in potential energy is negative of work done, then change in potential energy for this displacement dx is given by dU = -F(x)dx
In a three-dimensional reference frame, this equation is
dU = –\(\vec{F} \cdot d \vec{r}\) …… (1)
Conservation of Mechanical Energy of a Particle in a Conservative System
Suppose a conservative force \(\vec{F}\), acting on a particle of mass m moves, it from point A to point B.
If velocity of the particle at any point of its motion in the force field is v, then
\(\vec{F} \cdot d \vec{r}\) = m\(\frac{d \vec{v}}{d t} \cdot \vec{v} d t\) = \(m \vec{v} \cdot d \vec{v}\) = md\(\left(\frac{1}{2} v^2\right)\)
∴ Work done, to take the particle from A to B, by force \(\vec{F}\),
W = \(\int_A^B \vec{F} \cdot d \vec{r}\) = \(\left[\frac{1}{2} m v^2\right]_A^B\) = \(\frac{1}{2} m v_B^2\) – \(\frac{1}{2} m v_A^2\) ….. (1)
where vA and vB are velocities of the particle at A and B respectively.
From section 1.5 equation (1), we know
\(\vec{F} \cdot d \vec{r}\) = -dU
∴ \(\int_A^B \vec{F} \cdot d \vec{r}\) = \(-\int_A^B d U\) = UA – UB
= difference in potential energy between the points A and B …….. (2)
Comparing (1) and (2), UA – UB = \(\frac{1}{2} m v_B^2\) – \(\frac{1}{2} m v_A^2\)…… (3)
Equation (3) establishes that in a conservative field when kinetic energy of a particle increases, its potential energy decreases.
∴ \(\frac{1}{2} m v_A^2\) + UA = \(\frac{1}{2} m v_B^2\) + UB …….. (4)
If KA and KB are the kinetic energies of the particle at A and B respectively, then
KA + KA = KB + UB or, K + U = constant (E).
Hence in a conservative field i.e., [n a conservative system (where no dissipation of energy occurs due to forces like friction etc.), at each point sum of potential energy and kinetic energy remains constant. In other words total mechanical energy (E) of a particle In a conservative system remains constant.
This is the statement of the principle of conservation of mechanical energy
Total mechanical energy in a free fall under gravity remains constant: Let a body of mass m be at rest at point A [Fig.].
DE, the earth’s surface is taken as the reference plane and the height of point A from DE is h. As the body is at rest it has potential energy only and no kinetic energy.
Potential energy at A = mgh
Kinetic energy at A = 0
Total mechanical energy at A = mgh + 0 = mgh
Now the body is released and it starts falling freely. When the body reaches a point B, it acquires a velocity v. At B, the body has both potential energy (due to its height) and kinetic energy (due to its motion). Let AB = x
∴ Potential energy at B = mg(h – x)
Kinetic energy at B = \(\frac{1}{2} m v^2\) = \(\frac{1}{2} m \cdot 2 g x\)
= mgx [∵ v2 = u2 + 2as and u = 0]
∴ Total mechanical energy at B
= mg(h – x) + mgx = mgh
= total energy at A.
Also at C, when the body is about to touch the reference plane, its potential energy becomes zero. If the velocity acquired at that point is V, then kinetic energy at C
= \(\frac{1}{2} m V^2\)mV2 = \(\frac{1}{2} m \cdot 2 g h\) [∵ V2 = 2gh]
= mgh
Total mechanical energy at C = mgh + 0 = mgh
∴ Total energy at A = total energy at B = total energy at C.
Therefore, the total mechanical energy (= potential energy + kinetic energy) of a freely falling body, under the action of gravity, remains conserved at all positions.
As the body touches the ground, both its potential energy and kinetic energy become zero; the entire mechanical energy transforms into heat, sound and other forms of energy.
Total mechanical energy of a body, falling under gravity along a frictionless inclined surface, remains constant: Let a body of mass m be at rest at a point A on a friction-less plane of inclination θ. Earth’s surface CD is taken as the reference plane [Fig.].
Let the height of point A from the reference plane, GA = h. Being at rest, the body has no kinetic energy at A.
Potential energy at A = mgh
Kinetic energy at A = 0
∴ Total mechanical energy at A = mgh + 0 = mgh
On releasing the body, it falls along the incline, and reaches a point B such that, AB = x. Let the velocity of the body at B be v.
Acceleration of the body along the incline, a = g sinθ.
∴ v2 = 0 + 2gsinθ ᐧ x = 2gxsinθ [as v2 = u2 + 2as]
∴ Kinetic energy at B = \(\frac{1}{2}\)mv2 = \(\frac{1}{2}\) ᐧ m ᐧ 2gx sinθ
= mgx sinθ
Perpendicular height of B from CD plane
= FB = GE = GA – EA = h – xsinθ
[∵ sinθ = \(\frac{E A}{A B}\) or, EA = ABsinθ = xsinθ]
∴ Potential energy at B = mg(h – x sinθ)
∴ Total mechanical energy at B
= mg(h – xsinθ) + mgx sinθ
= mgh = total energy at A
At the point C, i.e., where the body just touches the plane CD, it has no potential energy. If the velocity of the body at that moment is V, then
V2 = 0 + 2gsinθ ᐧ AC = 2gsinθ ᐧ AC
and hence, kinetic energy at C = \(\frac{1}{2}\) ᐧ 2g sinθ ᐧ AC
= mgh
[∵ sinθ = \(\frac{G A}{A C}\) = \(\frac{h}{A C}\) or, AC sinθ = h]
∴ Total energy at C = 0 + mgh = mgh
Thus, total mechanical energy at A = total mechanical energy at B = total mechanical energy at C.
Hence, the total mechanical energy of a body, moving along a frictionless inclined plane under gravity, is conserved.
Total mechanical energy of a hydrogen gas-filed balloon rising upwards remains constant: Let us assume that air and earth form a system. If the earth surface is taken as the plane of reference, when the balloon is at rest on the ground, both its potential and kinetic energy = 0, i.e., total mechanical energy = 0.
Now the balloon is released. Upthrust, T due to air on the balloon is more than the weight, mg of the balloon filed with gas and so the balloon rises up. Let the mass of the gas-filled balloon = m and mean resultant upward force on it
= F = T – mg.
∴ Acceleration of the balloon = \(\frac{F}{m}\). If the velocity of the balloon is v at a height h from the earth surface, then
v2 = 2\(\frac{F}{m}\)h [according to the formula v2 = u2 + 2as]
Hence, kinetic energy at h = \(\frac{1}{2} m v^2\) = \(\frac{1}{2} m \cdot 2 \frac{F}{m} h\) = Fh
We know that when a stone falls freely under gravity, the work done is negative. Similarly, when the balloon rises by itself (without the help of any external agent) under the effect of the mean upward resultant force F, the work done is also negative. Here, the upthrust is not applied by an external agent. So, at height h the change in potential energy = -Fh. As the initial potential energy of the balloon on the ground was zero, the total potential energy at a height h = -Fh.
∴ Total mechanical energy of the balloon at height h = potential energy + kinetic energy = – Fh + Fh = 0 = total energy of the balloon on the ground before its release.
Thus, the total mechanical energy remains conserved for a hydrogen gas-filled balloon when it is rising up. It is to be noted that, with the increase in height h, the kinetic energy increases but the potential energy decreases equally.
Actually, the expression mgh for potential energy should be modified for a balloon as (mg – T) h, where T is the upward thrust exerted on the balloon by the air surrounding it. For a hydrogen-filled balloon, upward thrust is greater than its weight, i.e., T > mg. So the potential energy (mg – T)h is negative; this negative value goes on increasing with increase in height h.
Another similar event is observed when a piece of wood is held at the bottom of a bucket full of water. When it is released, it floats up by itself. It can be shown that the total mechanical energy remains conserved. As the piece of wood rises upwards, its kinetic energy gradually increases and the potential energy decreases equally.
Numerical Examples
Example 1.
After falling from a height of 200 m, water flows horizontally with a certain velocity. Ignoring any energy dissipation, find the velocity of flow.
Solution:
Potential energy changes into kinetic energy during the free fall of water.
Let mass of water = m, height = h, final velocity = v
Here, at the height of 200 m
potential energy (P.E.)i = mgh and kinetic energy (K.E)i = 0
when water falls and flows horizontally,
potential energy (P.E.)f = 0 and kinetic energy (K.E.)f = \(\frac{1}{2} m v^2\)
∴ From conservation of mechanical energy we have,
∴ mgh = \(\frac{1}{2} m v^2\)
or, v = \(\sqrt{2 g h}\) = \(\sqrt{2 \times 9.8 \times 200}\)
= 62.61 m ᐧ s-1.
Example 2.
Mass of the bob of a simple pendulum is 10 g and the effective length is 13 cm. The bob is pulled 5 cm away from the vertical and then released. What will be the kinetic energy of the bob when it passes through the lowest point?
Solution:
From Fig. we observe that, energy of the bob at point B potential energy = mgh = 10 × 980 × AC. Here OA = OB = 13 cm and CB = 5 cm.
∴ OC = \(\sqrt{13^2-5^2}\)
= \(\sqrt{144}\) = 12 cm
∴ AC = OA – OC
= 13 – 12 = 1 cm.
∴ Potential energy of the bob at B
= 10 × 980 × 1 = 9800 erg
∴ Kinetic energy of the bob at the lowest point A potential energy of the bob at B = 9800 erg.
Example 3.
After a collision with an Ideal spring, a body of mass 8 g, moving with a constant velocity of 10 cm ᐧ s-1 comes to rest. Force constant of the spring is 200 dyn ᐧ cm-1. If the total kinetic energy of the body is spent in compressing the spring, find the compression.
Solution:
Here, the kinetic energy of the body transforms into the potential energy of the spring.
∴ \(\frac{1}{2} m v^2\) = \(\frac{1}{2} k x^2\), where x = compression of the spring
∴ x = \(\sqrt{\frac{m v^2}{k}}\) = \(\sqrt{\frac{8 \times 10^2}{200}}\) = 2 cm.
Example 4.
Effective length of a pendulum is 50 cm and the mass of the bob is 4 g. The bob is drawn to one side until the string is horizontal, and is then released. When the string makes an angIe 60° with the vertical, what is the velocity and the kinetic energy of the bob?
Solution:
OB is the horizontal position of the string [Fig.].
At B, total energy of the bob
= potential energy = mgh
= (4 × 980 × 50) erg
At C, total energy of the bob
= kinetic energy + potential energy
= kinetic energy + 4 × 980 × AD
[cos 60° = \(\frac{O D}{O C}\) or, OD = \(\frac{50}{2}\) = 25 cm]
∴AD = OA – OD = 50 – 25 = 25 cm]
According to the law of conservation of energy, energy of the bob at C = energy of the bob at B
or, ICE. of the bob at C + RE. of the bob at C = K.E. of the bob at B + RE. of the bob at B
or, \(\frac{1}{2} m v^2\) + 4 × 980 × 25 = 0 + 4 × 980 × 50
[v = velocity of bob at C]
or, \(\frac{1}{2} m v^2\) = 4 × 980 × 25 = 98000
∴ v2 = \(\frac{98000 \times 2}{4}\) or, v = \(\sqrt{49000}\) = 222.36 cm ᐧ s-1
Example 5.
Mass of the bob of a pendulum is 100g and the length of the string is 1 m. The bob is initially held in such a way that the string is horizontal. The bob is then released. Find the kinetic energy of the bob when the string makes an angle of
(i) 0° and
(ii) 30° with the vertical.
Solution:
When the string is horizontal, the height of the bob above its lowest position = 1 m = 100 cm [Fig.].
Energy of the bob at point P = potential energy of the bob
= mgh = 100 × 980 × 1oo
= 98 × 105 erg.
i) At 0° angle with the vertical, the string holds the bob at its lowermost position. Hence, energy at B = kinetic energy of the bob = initial potential energy = 98 × 105 erg.
ii) When the string makes a 30° angle with the vertical, the height of the bob from its lowermost position
= BD = BA – DA = 100 – AC cos30°
= 1oo(1 – \(\frac{\sqrt{3}}{2}\)) = 13.4 cm.
Potential energy of the bob at C
= 100 × 980 × 13.4 erg;
kinetic energy at this position
= initial potential energy – potential energy at C
= 98 × 105 – (100 × 980 × 13.4) = 8486800 erg.
Example 6.
A body of mass 1kg falls to the ground from the roof of a building 20 m high. Find its
(i) Initial potential energy,
(ii) velocity when it reaches the ground,
(iii) maximum kinetic energy and
(iv) kinetic and potential energies at a position 2 m above the earth’s surface.
Solution:
i) Initial potential energy of the body
= mgh = 1 × 9.8 × 20 = 196 J.
ii) Suppose the body touches the ground with velocity v. Potential energy at roof level = kinetic energy just before touching the ground.
∴ 196 = \(\frac{1}{2}\) × 1 × v2
∴ v2 = 392 and v = 19.8 m ᐧ s-1
iii) Maximum kinetic energy
= initial potential energy = 196 J
iv) Potential energy at a height of 2m
= 1 × 9.8 × 2 = 19.6 J
∴ Kinetic energy at that height of 2 m = decrease in initial potential energy = 196 – 19.6 = 176.4 J.
Example 7.
A pump lifts 200 L of water per minute through a height of 5 m,and ejects it through an orifice 2 cm in diameter. Find the velocity of efflux of water and the power of the pump.
Solution:
Volume of water lifted by the pump in is
= \(\frac{200}{60}\) = \(\frac{10}{3}\)L = \(\frac{10^4}{3}\) cm3
Mass of this volume of water = \(\frac{10^4}{3}\)g = \(\frac{10}{3}\) kg
∴ Increase in potential energy in 1 second
= \(\frac{10}{3}\) × 9.8 × 5 J ᐧ s-1 = 163.33 J
∴ Power of the pump to raise water up to the height of 5 m = 163.33 W.
If the velocity of efflux is v, then πr2 × v = \(\frac{10^4}{3}\)
or, v = \(\frac{10^4 \times 7}{3 \times 22 \times(1)^2}\) = 1061 cm ᐧ s-1 = 10.61 mᐧs-1
Kinetic energy of the amount of water thrown out per second
= \(\frac{1}{2}\) × \(\frac{10}{3}\) × (10.61)2 J = 187.62 J.
∴ Power of efflux = 187.62 W
∴ Total power of the pump = 163.33 + 187.62 = 350.95 W.
Example 8.
A body of mass 10kg is raised to a height of 10 m with an upward force of 196 N. Find the work done by the upward force and the work done against gravitation. Show that the total energy in this case is equal to the work done by the upward force. [g = 9.8m ᐧ s-2] [HS’02]
Solution:
Work done by the upward force,
W = force × displacement
= 196 N × 10 m = 1960 N ᐧ 1960 J.
Upward acceleration of the body in absence of gravity,
a \(=\frac{\text { upward force }}{\text { mass }}\) = \(\frac{196}{10}\) = 19.6 m ᐧ s-2
∴ Effective upward acceleration,
a = a’ – g = 19.6 – 9.8 = 9.8 m ᐧ s-2
If the body starts from rest, and attains a velocity v at the height of 10 m, then from v2 = u2 + 2as,
v2 = 2 × 9.8 × 10m2.s-2
∴ Kinetic energy of the body at this height,
K = \(\frac{1}{2} m v^2\) = \(\frac{1}{2}\) × 10 × 2 × 9.8 × 10 = 980J
Work done against gravitational force,
W’ = force due to gravity × displacement
= mass × acceleration due to gravity × displacement
= 10 × 9.8 × 10 = 980 J
This work done against gravitational pull gets stored as potential energy V of the body. Hence, V = 980 J.
∴ Total mechanical energy of the body at a height of 10 m
= V + K = 980 + 980 = 1960 J = W
So, a part of the work done by the upward force changes into kinetic energy of the body, and the other part transforms itself into the stored potential energy. Thus, the total energy and work done by the upward force are equal.
Example 9.
A body of mass 10 kg moving with a speed of 2.0 m ᐧ s-1 on a frictionless table strikes a mounted spring and comes to rest. If the force constant of the spring be 4 × 105N ᐧ m-1, then what will be the compression on the spring?
Solution:
The kinetic energy of the body is
E = \(\frac{1}{2} m v^2\), where m and u be the mass and the speed of the body respectively.
On striking the spring, the kinetic energy of the spring due to compression is completely converted into the potential energy of the spring. If the spring is compressed through a distance x then its potential energy is
U = \(\frac{1}{2} k x^2\)
∴ \(\frac{1}{2} m v^2\) = \(\frac{1}{2} k x^2\)
or, x = \(v \sqrt{\frac{m}{k}}\) = 2 × \(\sqrt{\frac{10}{4 \times 10^5}}\) = 10-2 m = 1 cm
Example 10.
Fig. shows two blocks of masses m1 = 3 kg and m2 = 5 kg, both moving towards right on a friction-less surface with speeds u1 = 10 m ᐧ s-1 and u2 = 4 m ᐧ s-1 respectively. To the back side of m2 an ideal spring of force constant 1000 N ᐧ m-1 is attached. Calculate the maximum compression of the spring when the blocks collide?
Solution:
m1 = 3 kg, m2 = 5 kg,
u1 = 10 m ᐧ s-1, u2 = 4 m ᐧ s-1
Force constant, k = 1000 N ᐧ m-1
Let v be the speed of the combination.
Using the law of conservation of linear momentum,
m1u1 + m2u2 = (m1 + m2)v
∴ v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\) = \(\frac{3 \times 10+5 \times 4}{3+5}\) = 6.25 m ᐧ s-1
Also, let x be the maximum compression of the spring when the blocks collide.
∴ From conservation of mechanical energy, we get