WBCHSE Class 11 Physics Dissipation Of Energy Notes

Dissipation Of Energy

According to the law of conservation of energy, energy cannot be destroyed. But during the transformation of energy, some energy may change into such a form that has no practical utility and cannot be recovered in any usable form. This is called the dissipation of energy.

Dissipation Of Energy Example: Energy has to be supplied to a machine to make it work. But the work output (i.e., energy) is generally less than the energy supplied. This is because a part of the supplied energy is used to overcome friction and other resistive forces and this part transforms into heat or sound energy that cannot be used for practical purposes and is lost forever.

This is the dissipation of energy. It does not mean the destruction of energy. It simply denotes the transformation of energy into unusable form, also called unavailable energy. Various methods are used to reduce this dissipation of energy. But this could not be minimized to zero yet.

Efficiency Of A Machine: The ratio between the work output of a machine and the energy supplied to it, is called its efficiency. In real life, efficiency is less than 1, and it is often expressed in percentage by multiplying the ratio by 100.

Therefore, the efficiency of a machine = \(\frac{\text { work output of the machine }}{\text { energy supplied (input) }} \times 100 \%\)

For example, 90% efficiency of a machine means that, if 100 units of energy is supplied to the machine, work done by it will be 90 units.

Work Done Against Friction: The total mechanical energy of a body, falling under gravity along a frictionless inclined plane remains conserved. But, a frictionless surface is an ideal one, and cannot be obtained in practice. A frictional force always acts against the motion, and some work has to be done by the body against this force. As a result, some energy is dissipated.

Work And Energy Workd Done Against Friction

Suppose a body of mass m begins to move from point A under gravity along a rough inclined plane towards C.

The height of point A above the reference plane CD is h. Hence, the potential energy of the body at A = mgh. The body is at rest so its kinetic energy is zero there. Thus, the total mechanical energy of the body at A = mgh + 0 = mgh.

Under the action of the component mg sin# of the weight, the body starts moving down along the incline. Then a frictional force, f = μR = μmg cosθ acts upwards on the body along the inclined plane, where μ = coefficient of friction.

Hence, the resultant downward force along the plane,

F = mg sinθ- μmg cosθ = mg(sinθ – μ cosθ)

the acceleration along the plane, a = \(\frac{F}{m}\) = g(sinθ-μcosθ)

If the velocity of the falling body at B is v, where AB = x, then v² = 2ax = 2xg(sinθ- μcosθ)

Kinetic energy at B = \(\frac{1}{2}\) mv²

= \(\frac{1}{2}\) m · 2xg(sinθ – μcosθ)

= mgx(sinθ-μcosθ)

The potential energy at B = mg · DE = mg(DA-EA) =mg(h- xsinθ)

Hence, total mechanical energy at B

= mg(h- xsinθ) + mgx( sinθ- μcosθ)

= mgh-μmg cosθ · x = mgh – fx…(1)

Equation (1) shows that the mechanical energy at B is less than that at A by fx, which is the work done against the frictional force to cover a distance x along the plane. This amount fx of energy is transformed into unavailable form, in order to overcome the frictional force against motion.

So, the total energy dissipated during the sliding of the body along an inclined plane of length l (=AC) = fl = μmgl cosθ.

The above discussions show that, in the presence of dissipative forces like friction, mechanical energy does not remain conserved for a system. We see that,

Total mechanical energy at A = total mechanical energy at B +fx

The work fx, done against friction, actually transforms into heat energy at the surface of contact of the body with the plane. This heat can never be recovered in any usable form. However, taking this heat into consideration, we see that the total energy is certainly conserved.

Unit 4 Work Energy Power Chapter 1 Work And Energy

Dissipation Of Energy Numerical Examples

Example 1. A block weighing 250 N Is pulled over a horizontal plane at a constant velocity up to a distance of 10 m. The coefficient of kinetic friction Is 0.2 and the force is applied by a string, attached to the block, Inclined at 60° with the vertical. Find the work done against friction.
Solution:

Let the force applied on the block be F

Horizontal component of the applied force along the plane

= \(F \sin 60^{\circ}=\frac{\sqrt{3}}{2} F\) and its vertical component

= \(F \cos 60^{\circ}=\frac{F}{2} .\)

Work And Energy A Block Weigthing Is Pulled Over A Horizontal Plane At A Constant Velocity

Since there is no vertical acceleration of the block, net force acting vertically is zero.

i.e., R + Fcos60° = W (where R is the normal reaction on the block)

∴ R = W – \(\frac{F}{2}\)

As the body is moving with a uniform velocity, the horizontal component of applied force = frictional force

or, \(\frac{\sqrt{3}}{2} F=\mu R=\mu\left(W-F \cos 60^{\circ}\right)=0.2\left(250-\frac{F}{2}\right)\) = 50-0.1 F

∴ F = \(\frac{50}{0.866+0.1}=\frac{50}{0.966} \mathrm{~N}\)

Hence, work done by the applied force

= \(F \sin 60^{\circ} \times 10=\frac{50}{0.966} \times \frac{\sqrt{3}}{2} \times 10=448.25 \mathrm{~J} .\)

Example 2. A particle Is sliding down along an inclined plane. The frictional force is 0.2 times the normal reaction, and the inclination of the plane is 60J. What Is the acceleration of the particle? If the mass of the particle is 1 g, find the change in the sum of potential and kinetic energies of the particle as it slides down the plane by 1 m.
Solution:

Let the acceleration of the particle along the inclined plane = a and the downward force on the particle along the plane = mg sinθ – f

Work And Energy A particle Slidding Down Along A Inclined Plane

∴ ma = mgsinθ – μR = mg sinθ – μmg cosθ

or, a = g (sinθ – nμcosθ) =9.8 (sin60° – 0.2cos60°)

= \(9.8\left(\frac{\sqrt{3}}{2}-0.2 \times \frac{1}{2}\right)=7.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Change in mechanical energy

= work done against friction = μmg cosθ · s

= 0.2 x 0.001 x 9.8 x \(\frac{1}{2}\) x 1 = 0.00098 J.

Example 3. A box of mass 12 kg is pushed up by a distance of 10 m on the application of a 100 N force along a plane of inclination 30°. If the coefficient of friction is \(\frac{1}{\sqrt{3}}\), find the work done against friction, [g = 10 m · s-2]
Solution:

Frictional force, f = μR = μmg cos30°

Work And Energy A Box Of Mass Is Pushed Up By A Distance Of Inclination

Hence work done against friction, W = fs

= μmg cos30° · s

= \(\frac{1}{\sqrt{3}} \times 12 \times 10 \times \frac{\sqrt{3}}{2} \times 10=600 \mathrm{~J}\)

Example 4. A car of mass 1000 kg moves up at 40 km · h-1 along an inclined plane of slope \(\frac{1}{50}\). The coefficient of rolling friction between the road and the wheels of the car is 0.3. Find the power of the car engine.
Solution:

The angle of inclination = θ.

∴ Slope = tanθ = \(\frac{1}{50}\) ≈ sinθ [θ is small]

Work And Energy A Car Of Mass 1000kgs Moving Up At Inclined Plane

The resultant downward force on the car along the incline

= mg sinθ + f

= mg sinθ + μR

= mg sinθ + μmg cosθ = mg(sinθ + μcosθ)

= \(1000 \times 9.8\left(\frac{1}{50}+0.3 \times 1\right)\) (as is very small cosθ ≅ 1)

= 9800 x 0.32 = 3136 N

Velocity of the car =40 km · h-1 = 11.11 m · s-1

∴ Power of the car =3136 x 11.11 N · m · s-1 =34840.96 W = 34.84 kW.

Example 5. An engine, working at a constant rate, is pulling a train of mass 500 tonne along a plane of inclination sin-1\(\frac{1}{100}\).frictional force per metric tonne is 49 N and the train is moving with a velocity of 10 m · s-1, what is the power of the engine in kilowatt? [1 tonne (metric ton) = 1000 kg]
Solution:

Downward effective force on the train along the inclined plane = mg sinθ + frictional force (f)

= 500 x 1000 x 9.8 x \(\frac{1}{100}\) + 500 x 49 = 7500 x 9.8 N

Velocity of the train = 10 m · s-1

Since the car is moving upward with a constant velocity, the force applied by the car’s engine, F must exactly balance the net downward force.

∴ F = 3136 N

Hence, power of the engine = 7500 x 9.8 x 10 = 735000 J · s-1 = 735000 W = 735 kW.

Example 6. A loaded lorry of total mass 5000 kg can come down from the top of a slope (1:40) effortlessly at 18 km · h-1. What should be the horsepower of its engine so that it can go up with the same speed, from the base to the top? Resistance due to friction may be taken to be the same in both cases.
Solution:

Velocity of the lorry = \(18 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{18 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) \(\sin \theta=\frac{1}{40}\)

As the speed of the lorry remains constant for the downward journey, the friction just balances the downward active force mg sinθ.

∴ Frictional force, f = mg sinθ.

Effective force on the lorry for its upward journey = mg sinθ + f = mg sinθ + mg sinθ = 2mg sinθ

∴ Power of the engine = effective force on the lorry x velocity of the lorry

= 2mgsinθ x 5 = 2 x 5000 x 9.8 x \(\frac{1}{40}\) x 5

= 12250 W = \(\frac{12250}{746}\)hp = 16.42 hp.

Example 7. A car of mass 500 kg is moving up along an inclined surface of slope \(\frac{1}{25}\) at a constant speed of 72 km · h-1. If the coefficient of friction between the road and the car wheel is 0.1, find the power of the car engine (g = 9.8 m · s-2).
Solution:

Frictional force, f = μmg cosθ, v = 72 km · h-1 = 20 m · s-1, m = 500 kg, g = 9.8m · s-2, μ = 0.1 and tanθ = \(\frac{1}{25}\)

∴ sinθ  = \(\frac{1}{\sqrt{626}}\) and

cosθ = \(\frac{25}{\sqrt{626}}\)

Work And Energy A Car Of Mass Is Moving Up Along An Inclined Surface

F = force opposing the motion of the car = f + mg sinθ = mg(μcosθ + sinθ)

∴ Power of the car engine, P = Fv= mg(μcosθ + sinθ)v

= \(500 \times 9.8 \times\left(\frac{0.1 \times 25}{\sqrt{626}}+\frac{1}{\sqrt{626}}\right) \times 20\)

= \(13709 \mathrm{~J} \cdot \mathrm{s}^{-1}=13709 \mathrm{~W} .\)

Example 8. A 1.5 m long chain of mass 0.8 kg is kept on a horizontal table and a part of its length hangs from the edge of the table. When the length of the hanging part is one-third the total length of the chain, it starts sliding off the table. What will be the work done by friction when the whole length of the chain slides off the table?
Solution:

When \(\frac{1}{3}\) of the chain is hanging, it starts sliding.

In this condition, frictional force = weight of the hanging part of the chain

or, μ x \(\frac{2}{3}\)mlg= \(\frac{1}{3}\)mlg [μ = coefficient of friction, m = mass per unit length of the chain, and l = length of the chain]

∴ μ = \(\frac{1}{2}\) = 0.5

Work And Energy A Long Chain Of Mass Kept On A Horizontal Table

When the whole length of the chain slides off the table, the effective frictional force on the chain = 0.

∴ Effective average frictional force on the chain = \(\frac{\frac{2}{3} \mu \mathrm{mlg}+0}{2}=\frac{1}{3} \mu \mathrm{mlg} .\)

The chain moves through a distance of \(\frac{2}{3}\). l against the effective friction. Work done against friction is, therefore,

W = \(\frac{1}{3} \mu m g l \times \frac{2}{3} l=\frac{2}{9} \mu m g l^2\)

= \(\frac{2}{9} \times 0.5 \times \frac{0.8}{1.5} \times 9.8 \times(1.5)^2\)

= \(1.3 \mathrm{~J} .\)

Example 9. A body of mass 10 kg is pushed up 50 cm from the ground, along a plane inclined at 45° to the horizontal. if the coefficient of friction is 0.2, then calculate the work done.
Solution:

Here, h = 50 cm = 0.5 m, m = 10 kg, g = 9.8 m · s-2, θ = 45°, μ = 0.2

Let the friction acting on the body be f. Then, f = μmg cosθ

The force against which the body is pushed up is F = f + mg sinθ

= mg(μ cosθ + sinθ)

The body is pushed up by a distance \(\frac{h}{\sin \theta}\) along the inclined plane.

Therefore, the work done is, W = \(\frac{F h}{\sin \theta}=\frac{m g h}{\sin \theta}(\mu \cos \theta+\sin \theta)\)

= \(\frac{10 \times 9.8 \times 0.5}{\sin 45^{\circ}} \times\left(0.2 \times \cos 45^{\circ}+\sin 45^{\circ}\right)\)

= \(49 \times \sqrt{2} \times\left(0.2 \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=58.8 \mathrm{~J} .\)

Example 10. By application of a force F, a body of mass m Is raised to the top of a hill. F is tangential along the whole path. If the height of the hill Is h, the length of the base of the hill is l, and the coefficient of friction is μ, then find the work done.
Solution:

The total path from the bottom to the top of the hill can be considered as an assembly of a large number of inclined planes of varying angles of inclination. Consider an incline of length Δs.

Its vertical height is Δh and the angle of inclination is α.

Total work done in lifting the body along this inclined surface = work done against gravity + work done against friction.

Work And Energy By Application Of Force Is Raised By Hill

∴ \(\Delta W=m g \Delta h+\mu m g \cos \alpha \cdot \Delta s\)

= \(m g \Delta h+\mu m g \cos \alpha \frac{\Delta l}{\cos \alpha}=m g(\Delta h+\mu \Delta l)\)

∴ Total work done to lift the body up to the top of the hill,

W = \(\sum \Delta W=m g\left(\sum \Delta h+\mu \sum \Delta l\right)\)

= \(m g(h+\mu l)\)

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