Work, Energy And Power

Work done by a force f on a object which give body a displacement of d

where,
w is the work done on the object by force f
f is the force applied to the object
d is the displacement of the object after application of force
θ is the angle between force and displacement


A. The displacement of the bucket and the force applied on the bucket by the man have the same direction i.e. θ=0. Hence, the work done by a man is positive and the work is done on the bucket.

⇒ (∵ cos 0⁰=1)

Hence, work done is positive.

B. The gravitational force acts downwards on the bucket. But the displacement of the bucket is upwards i.e. θ=180. Hence, the work done is negative and θ=180

(∵ cos 180⁰=-1)

Hence, work done is negative.

C. Frictional force always acts opposite to the direction of motion of object i.e. θ=180. Hence, work done by frictional force is always negative.

(∵ cos 180⁰=-1)

Hence, work done is negative.

D. In this case, frictional force acts on the body due to the roughness of the horizontal plane in a direction opposite to the direction of motion. To obtain a uniform velocity, a uniform force must be applied in the direction of motion i.e. θ =0. Hence, the work done by the force is positive.

⇒ (∵ cos 0⁰=1)

Hence, work done is positive.

E. The resistive force always acts in the direction opposite to that of the motion of pendulum to bring the pendulum to rest i.e. θ =180. Hence, the work done by the resistive force is negative.

(∵ cos 180⁰=-1)

Hence, work done is negative.

Given,

Mass of the body, m = 2kg

Force applied, F = 7N

Coefficient of kinetic friction, μ = 0.1

A. Time, t = 10s

Using Newton’s second law of motion,

F=ma

where, a is the acceleration caused by force F on a body of mass m

∴

⇒ a=3.5 m/s²

Frictional force is given by

f=μmg

Where, μ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

∴ f=0.1 × 2kg × 9.8ms^-2

⇒ f=1.96N

Acceleration due to friction is given by

a_f=f/m

∴ (The negative sign is due to the fact that the frictional force acts opposite to the direction of motion)

⇒ a_f=-0.98ms^-2

Total acceleration = acceleration due to applied force + acceleration due to friction

Total acceleration, a=a_F+a_f

⇒ a=3.5ms^-2+(-0.98ms^-2)

∴ a=2.52ms^-2

The displacement of the object is given by Newton’s second equation of motion as

⇒ s= 0 + (1/2)(2.52ms^-2)(10s)²

⇒ s=126 m

Work done by the applied force is given by

W=F∙s

⇒ W=7 N × 126 m

⇒ W=882 J

B. Work done by friction is given by

W=f∙s

⇒ W= (-1.96 N) × (126 m)

⇒ W= -246.96 J

C. Net force on the body, F_t=F+f

∴ F_t=7 N + (-1.96 N) = 5.04 N

Work done by total force is given by

W_t=F_t∙s

⇒ W_t=5.04 N × 126 m

⇒ W_t=635.04 J

D. Using newton’s First equation of motion, the final velocity is calculated as

v=u+at

⇒ v=0+(2.52 ms^-2)(10 s)

⇒ v=25.2 m/s

Change in kinetic energy,

So, ΔK = ()(2kg)(25.2ms^-1)² – ()(2kg)(0)

⇒ ΔK = 635.04 J

This result is in accordance with Work-Energy theorem which states that the change in kinetic energy of an object is equal to the net work done on the object.

The kinetic energy of a particle of mass m moving with a velocity v is given as

K=

Since m is always positive, K is also always positive.

The total energy (E) of a particle is equal to the sum of Kinetic energy (K) and Potential energy (PE).

⇒ E = K + PE

⇒ K = E – PE

a) For x>a, PE(=V₀)>E.

This means that the kinetic energy is negative. Hence, the particle cannot exist in the region x>a.

b) For x<a and x>b, PE(=V₀)>E.

This means that the kinetic energy is negative. Hence, the particle cannot exist in the region x<a and x>b.

c) In the region a<x<b, the kinetic energy is positive. The potential energy is -V₁. So, K = E-(-V₁) = E + V₁.

In order for K to be positive, E> -V₁. So, the minimum total energy of the particle is -V₁.

d) For the regions -b/2<x<-a/2 and a/2<x<b/2, the potential energy of the particle is greater than the total energy which suggests that the kinetic energy is negative in this region. Hence, the particle cannot exist in this region.

The lowest potential energy is -V₁. So, K = E-(-V₁) = E+V₁.

For K to be positive, E should be greater than -V₁. So, minimum total energy is -V₁.

Given,

Total energy of the particle, E = 1 J

Force constant, k = 0.5 N m^-1

The total energy is equal to the sum of kinetic energy and potential energy.

So, E = PE + K

⇒ E = kx² + mv²

So, 1 = kx² + mv²

When the velocity of the particle is zero i.e., at the turning point, the kinetic energy is zero.

∴ 1 = kx²

⇒ x² = 2/k

⇒ x² = 2/0.5

⇒ x² = 4

⇒ x = �2

Hence, the particle turns back at x= �2.

A. The heat energy required for burning is obtained at the expense of the rocket. The mass of the rocket is consumed to supply for the loss in total energy of the rocket.

B. Gravitational force is conservative in nature. The work done by a conservative force in a closed cycle is zero. Hence, work done by the gravitational attraction of sun is zero.

C. When the satellite comes closer to the earth, its potential energy decreases (∵ PE = mgh where h is the height of the object from the earth). To maintain constant total energy, the kinetic energy increases and hence velocity increases.

D. Given,

Mass = 15 kg

Displacement, s = 2 m

In figure 1:

Work done, W = F∙s

⇒ W = F×s×cosθ

⇒ W = F×s× cos 90°

⇒ W = 0 (∵ cos 90° = 0)

In figure 2:

Work done, W = F∙s

⇒ W = F s cosθ

⇒ W = m g s cos 0°

⇒ W = 15 kg ×9.8 m s^-2× 2 m × 1

⇒ W = 294 J

Hence, more work is done in figure 2.

A. When a conservative force does positive work on a body, the potential energy of the body decreases.

Explanation: When a conservative force does positive work on a body, the object is displaced in the direction of force and towards the centre of the force such that the distance between the source and the object decreases. So, potential energy also decreases.

B. Work done by a body against friction always results in a loss of its kinetic energy.

Explanation: Friction acts always in the direction opposite to that of motion and results in a decrease in the velocity of the body. Hence, friction decreases the kinetic energy of the body.

C. The rate of change of total momentum of a many-particle system is proportional to the external force on the system.

Explanation: Internal forces cannot change the total energy of a system in order to maintain a constant energy of the system. Hence, only external forces can bring about a change in the total momentum of a many-particle system.

D. In an inelastic collision of two bodies, the quantities which do not change after the collision are total linear momentum of the system of two bodies.

Explanation: The total momentum of a system remains constant in a collision irrespective of the nature of the collision according to law of conservation of linear momentum.

A. False

Explanation: In an elastic collision, the energy and momentum of the entire system is conserved and not of individual bodies.

B. False

Explanation: External forces can do work on the system and hence, they can change the total energy of the system.

C. False

Explanation: Work done in the motion of a body over a closed loop is zero only for conservative forces.

D. True

Explanation: In inelastic collisions, the final kinetic energy is always less than the initial kinetic energy because some of the energy is lost during the process in the form of sound, heat, etc.

A. In an elastic collision of two billiard balls, the total kinetic energy is not conserved during the short time of collision of the balls. During this period, the kinetic energy of the balls is zero. All the energy is stored in the form of potential energy.

B. The total linear momentum is always conserved during the entire duration of collision. The individual linear momenta change but the total linear momentum always remains constant.

C. For an inelastic collision, kinetic energy is not conserved but the total linear momentum is conserved.

D. The collision is elastic in nature. The forces involved are conservative in nature since potential energy depends on the separation of the centres.

NOTE: Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

The correct answer is (ii).

Explanation:

From Newton’s first equation of motion,

v = u + at

where, v is the final velocity

u is the initial velocity

a is the acceleration of the body

t is the time taken

Given, u = 0.

So, v = at

Power is given by

P = F × v

⇒ P = ma × at

⇒ P = ma²t

Since m and a are constant,

P ∝ t

The correct answer is (iii).

Explanation:

Power, P = F v = k (constant)

Considering the dimensions,

[P] = [F][v]

⇒ [P] = [MLT^-2][LT^-1]

⇒ [P] = [ML²T^-3]

So, [ML²T^-3] = k

⇒ [L²T^-3] = k

⇒ [L²/T³] = k

⇒ [L²] = k[T³]

⇒ [L²] ∝ [T³]

⇒ [L] ∝ [T^3/2]

Hence, l ∝ t^3/2

Given,

Force, N

Displacement, m

Work done, W =

So, W =

⇒ W = 3×4

⇒ W = 12 J

For an electron:

Mass of the electron, m_e= 9.11×10^-31kg

Let v_e be the velocity of the electron.

Kinetic energy of the electron, K_e = m_ev_e²

⇒ 10 = m_ev_e²

⇒ m_ev_e²=20 ……………….(1)

For a proton:

Mass of the proton, m_p= 1.67×10^-27kg

Let v_p be the velocity of the proton.

Kinetic energy of the proton, K_p=m_pv_p²

⇒ 100 = m_pv_p²

⇒ m_pv_p²=200 ……………………(2)

Dividing equation (1) by (2),

⇒

⇒

⇒

⇒

⇒ = 13.54

∴ v_e:v_p = 13.54 : 1

Given,

Radius of the rain drop, r = 2 mm = 2×10^-3 m

Total displacement of the rain drop, s = 500 m

Assuming the rain drop to be spherical,

Volume of the rain drop, V = (4/3)πr³

∴ V = (4/3)×π×(2×10^-3)³

⇒ V = 3.35×10^-8 m³

We know, Density of water, ρ = 10³ kg m^-3

So, Mass of rain drop, m = ρV

⇒ m = 10³ kg m^-3 × 3.35×10^-8 m³

⇒ m = 3.35×10^-5 kg

Gravitational force on the rain drop, F = mg

⇒ F = 3.35×10^-5 kg × 9.8 m s^-2

⇒ F = 3.28×10^-4 N

Work done by the gravitational force in the first half of the motion, W₁=F(s/2)

⇒ W₁ = 3.28×10^-4× (500/2)

⇒ W₁ = 8.21×10^-2 J

The work done by the gravitational force is equal in both the halves of the motion.

So, work done by the gravitational force during second half of motion, W₂ = 8.21×10^-2 J

If no resistive force is present, then the total energy of the rain drop will remain conserved according to law of conservation of energy.

Total energy at the starting point, E = mgh

⇒ E = 3.35×10^-5 kg × 9.8 m s^-2 × 500 m

⇒ E = 0.164 J

Due to the resistive force, some energy will be lost. The drop hits ground with velocity 10 m s^-1.

Total energy of the drop when it hits the ground, E’ = (1/2)mv²

⇒ E’ = (1/2) × 3.35×10^-5 kg × (10 m s^-1)²

⇒ E’ = 1.675×10^-3 J

So, loss of energy due to resistive force, ΔE = E’-E

⇒ ΔE = 1.675×10^-3 J – 0.164 J

⇒ ΔE = -0.162 J

NOTE: The negative sign indicates that energy is lost due to the resistive forces.

The linear momentum is always conserved irrespective of the nature of collision. Since the gas molecule rebounds with the same speed, the rebound velocity of the wall is zero. Hence, the kinetic energy is conserved as there is no change in velocities. Thus, this is an elastic collision.

NOTE: Initial and final velocities of the molecule and wall remain same before and after the collision. Hence, the kinetic energies also remain same.

Given,

Volume of the tank, V = 30 m³

Operation time of pump, t = 15 min = 15×60 sec = 900 sec

Height of the tank from ground, h = 40 m

Efficiency of the pump, η = 30%

We know, density of water, ρ = 10³ kg m^-3

So, Mass of water, m = ρV

⇒ m = 10³ m³ × 30 kg m^-3

⇒ m = 3 × 10⁴ kg

Output power, P_o = Work done/Time of operation

∴ P_o = mgh/t

⇒ P_o = (3×10⁴ kg × 9.8 m s^-2 × 40 m)/(900 sec)

⇒ P_o = 1.307×10⁴ W = 13.07 kW

We know,

Efficiency (η) = Output power (P_o)/Input power (P_i)

∴ Input power (P_i)= Output power (Po)/Efficiency (η)

⇒ P_i = (1.307×10⁴ W)/(0.3)

⇒ P_i = 4.355×10⁴ W = 43.55 kW

From the figure, it is observed that the linear momentum is conserved in all the cases. For an elastic collision, kinetic energy is also conserved.

Let us assume that the mass of each ball is m.

Total kinetic energy before collision,

K_i = (1/2)mV² + (1/2)m(0)² + (1/2)m(0)²

⇒ K_i = (1/2)mV²

Case (i):

Total kinetic energy after collision,

K_f = (1/2)m(0)² + (1/2)m(V/2)² + (1/2)m(V/2)²

⇒ K_f = (1/8)mV² + (1/8)mV²

⇒ K_f = (1/4)mV²

The total kinetic energy of the system before collision is not equal to the total kinetic energy after collision in this case. The total kinetic energy is not conserved. Hence, the collision is not elastic.

Case (ii):

Total kinetic energy after collision,

K_f = (1/2)m(0)² + (1/2)m(0)² + (1/2)mV²

⇒ K_f = (1/2)mV²

The total kinetic energy of the system before collision is equal to the total kinetic energy after collision in this case. The total kinetic energy is conserved. Hence, the collision is elastic.

Case (iii):

Total kinetic energy after collision,

K_f = (1/2)m(V/3)² + (1/2)m(V/3)² + (1/2)m(V/3)²

⇒ K_f = (1/18)mV² + (1/18)mV² + (1/18)mV²

⇒ K_f = (1/6)mV²

The total kinetic energy of the system before collision is not equal to the total kinetic energy after collision in this case. The total kinetic energy is not conserved. Hence, the collision is not elastic.

Thus, only case (ii) is an elastic collision.

The bob A will not rise after the collision. It will stay at rest because the collision is elastic. In an elastic collision, the velocities are interchanged. Hence, the bob A will come to rest and the bob B will move with the velocity of bob A after collision.

NOTE: Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.

Given,

Length of the pendulum, l = 1.5 m

Energy dissipated against air resistance, E = 5%

Let m be the mass of the bob and v be its velocity at the lowermost point.

At the extreme end:

Potential energy of the bob, P = mgl

Kinetic energy of the bob, K = 0 (∵ bob is at rest)

Total energy of bob, E = P + E

⇒ E = mgl ……………(1)

At the lowermost point:

Potential energy of the bob, P = 0

Kinetic energy of the bob, K = (1/2)mv²

Total energy of the bob, E = P + K

⇒ E = (1/2)mv²

During the motion from an extreme end to the lowermost point, the bob loses 5% of its energy, i.e., it has 95% of the energy.

So, (1/2)mv² = (95/100)mgl

⇒ v² = 95gl/50

⇒ v =

⇒ v =

⇒ v = 5.28 m s^-1

The trolley, at first, is moving with a constant velocity. Hence, there is no net external force acting on the trolley. Even if the sand starts leaking, there is no external force on the trolley. Hence, the speed of the trolley will not change.

Given,

Mass of the body, m = 0.5 kg

Velocity of the body is given by

v =a x^3/2 where a = 5 m^–1/2 s^–1

At x = 0, initial velocity, u = 0

At x = 2, final velocity, v = 5 (2)^3/2 = m s^-1

According to work-energy theorem, work done is equal to the change in kinetic energy of the body.

Work done, W = ΔK

⇒ W = (1/2)mv² – (1/2)mu²

⇒ W = (1/2) × 0.5 kg × ( m s^-1)² – (1/2) × 0.5 kg × 0²

⇒ W = 50 J

Given,

Area swept by the blades = A

Velocity of wind flow = v

Density of air = ρ

(a) Volume of the wind flowing through the windmill per second, V = Av

Mass of the wind flowing through the mill per second,

m = ρAv

Mass of the wind flowing through the mill in time t,

M = ρAvt

(b) Kinetic energy of air, K = (1/2)Mv²

⇒ K = (1/2)(ρAvt)v²

⇒ K = (1/2)ρAv³t

(c) Given,

Area swept by the blades, A = 30 m²

Velocity of wind flow, v = 36 km/h

⇒ v = (36×1000 m)/(60×60 sec)

⇒ v = 10 m/s

Density of air, ρ = 1.2 kg m^-3

Electric energy produced, E = 25% of wind energy

⇒ E = (25/100)(1/2) ρAv³t

⇒ E = (25/100)(1/2)× 1.2 kg m^-3 × 30 m² × (10 m s^-1)³ × t

⇒ E = 4500t J

Electric power, P = Electric energy (E)/ Time (t)

⇒ P = 4500t/t

⇒ P = 4500 W = 4.5 kW

(a) Given,

Mass of the weight, m = 10 kg

Height upto which the weight is lifted, h = 0.5 m

Number of times the weight is lifted, n = 1000

Work done against the gravitational force,

W = nmgh

⇒ W = 1000 × 10 kg × 9.8 m s^-2 × 0.5

⇒ W = 49000 J = 49 kJ

(b) 1 kg of fat is equivalent to 3.8 × 10⁷ J.

Efficiency of conversion, η = 20%

Mechanical energy supplied by body of the person,

E = (20/100)×3.8×10⁷ J

⇒ E = 7.6 × 10⁶ J

Equivalent mass of fat lost by dieter,

M =

⇒ M = 6.45×10^-3 kg = 6.45 g

Given,

Power used by the family, P = 8 kW = 8000 W

Solar power incident per square meter, E = 200 W

Efficiency of energy conversion, η = 20%

(a) Let A be the area required to supply 8 kW power.

So, 8×10³ = 20% × (A × 200)

⇒

⇒ A = 200 m²

(b) 200 m² is equivalent to a rooftop with dimensions 14m×14m.

Given,

Mass of the bullet, m = 0.012 kg

Horizontal speed (initial speed) of bullet, u = 70 m/s

Mass of wood block, M = 0.4 kg

Initial speed of the block, u’ = 0

After collision, the bullet sticks to the block. Let the final velocity of bullet and block after collision be v.

According to law of conservation of linear momentum,

mu + Mu’ = (m+M)v

⇒

⇒

⇒ v = 2.039 m/s

For the bullet-wooden block system,

Mass of the system, m_t = 0.012 kg + 0.4 kg = 0.412 kg

Velocity of the system, v = 2.039 m/s

Applying the law of conservation of energy, the potential energy at the highest point is equal to the kinetic energy at the lowest point.

Hence, m_tgh = (1/2)m_tv²

⇒ h = v²/2g

⇒ h = (2.039 ms^-1)²/(2×9.8 ms^-2)

⇒ h = 0.212m

So, the block will rise to a height of 0.212 m or 21.2 cm.

Heat produced (H) = Kinetic energy of the bullet – Kinetic energy of the system

⇒ H = (1/2)mu² – (1/2)m_tv²

⇒ H=(1/2)×0.012 kg×(70 ms^-1)² – (1/2)×0.412 kg×(2.039 ms^-1)²

⇒ H = 29.4 J – 0.856 J

⇒ H = 28.54 J

It is given that AB and AC are two frictionless planes inclined to the horizontal at angles θ₁ and θ₂ respectively.

The kinetic energy at B will be equal to the kinetic energy at C which will be the potential energy at A due to law of conservation of energy since there are no losses.

Let v₁ and v₂ be the final velocities of the stones on the planes AB and AC respectively and m be the mass of each stone.

Then, mgh = (1/2)mv₁² = (1/2)mv₂²

⇒ v₁ = v₂ = v (say)

If a₁ and a₂ are the accelerations of the two stones along the planes AB and AC respectively, then

a₁ = gsinθ₁

a₂ = gsinθ₂

As θ₂>θ₁, a₂>a₁.

Initial velocities of the two stones are zero.

From Newton’s first equation of motion,

v = u + at

⇒ v = 0 + at

⇒ v = at

⇒ t = v/a

Since v is constant for both the stones,

t ∝ 1/a

Since a₂>a₁, t₂<t₁.

So, the stone on plane AC will take less time than the stone on plane AB to reach the bottom.

Given,

Mass of the block, m = 1 kg

Spring constant, k = 100 N m^-1

Displacement of the block, s = 10 cm = 0.1 m

The forces can be shown in the figure as follows:

At equilibrium,

Surface normal reaction force, N = mgcos37°

Frictional force, f = μN where, μ is the coefficient of friction

⇒ f = mgsin37°

Net force acting on the block, F = mgsin37° - f

⇒ F = mgsin37° - μmgcos37°

⇒ F = mg(sin37° - μcos37°)

At equilibrium, work done by the block is equal to the potential energy of the spring according to law of conservation of energy.

So, mg(sin37° - μcos37°)s = (1/2)ks²

⇒ 1 kg×9.8 m s^-2×(0.602 – μ0.799)×0.1 = (1/2)×100Nm^-1×(0.1)²

⇒ 0.602 - μ × 0.799 = 0.51

⇒ μ × 0.799 = 0.092

⇒ μ = 0.115

Given,

Mass of the bolt, m = 0.3 kg

The speed of the bolt, v = 7 m s^-1

Length of the elevator, h = 3 m

The relative velocity of the bolt with respect to the elevator is zero. So, the potential energy gets converted into heat energy upon impact according to law of conservation of energy.

Heat produced, H = Loss of potential energy

⇒ H = mgh
where
m is the mass of the bolt
g is the acceleration due to gravity
h is the height of the elevator

⇒ H = 0.3 kg × 9.8 m s^-2 × 3 m

⇒ H = 8.82 J
The answer will not change because of the fact that the relative velocity of the bolt with reference to the elevator is 0.

Given,

Mass of the trolley, m = 200 kg

Speed of the trolley, v=36km/h=(36×1000 m)/(60×60 s)= 10m/s

Mass of the child, m’ = 20 kg

Initial momentum of the system involving the child and the trolley, p_i = (m+m’)v

⇒ p_i = (200 kg + 20 kg)×10 ms^-1

⇒ p_i = 2200 kg m s^-1

Let v’ be the final velocity of the trolley with respect to the ground.

Final velocity of the child with respect to ground, v’’ = Final velocity of trolley with respect to ground – final velocity of child with respect to trolley.

So, v’’ = v’ – 4

Final momentum of the system, p_f = mv’ + m’v’’

⇒ p_f = mv’ + m’(v’-4)

⇒ p_f = 200v’ + 20(v’-4)

⇒ p_f = 220v’-80

According to the law of conservation of linear momentum,

Initial momentum (p_i) of system = Final momentum (p_f) of system

⇒ 2200 = 220v’-80

⇒ 220v’ = 2280

⇒ v’ = 2280/220

⇒ v’ = 10.36 m/s

Length of the trolley, l = 10 m

Speed of the child with respect to trolley, v’ = 4m/s

Time taken by the boy to cover the distance, t = l/v’

⇒ t = 10m/4 ms^-1

⇒ t = 2.5 s

Distance moved by the trolley in this time, s = v’t

⇒ s = 10.36 ms^-1 × 2.5 s

⇒ s = 25.9 m

The potential energy of a system containing two masses is inversely proportional to the distance between the two masses. Also, if R is the radius of each billiard ball and r is the separation between them, then the potential energy becomes zero when the two balls touch each other, i.e., r = 2R. Only (v) satisfies these conditions. Hence, (i), (ii), (iii), (iv) and (vi) cannot possibly describe a collision of two billiard balls.

Let the masses of electron and proton be m and m’ respectively and their velocities be v and v’ respectively.

Using law of conservation of linear momentum,

Momentum of electron + momentum of proton = momentum of neutron

mv + m’v’ = 0

⇒ v’ = -mv/m’

This shows that the electrons and protons move in the opposite directions. If mass Δm has been converted into energy, then

(1/2)mv² + (1/2)m’v’² = Δmc²

⇒ (1/2)mv² + (1/2)m’(-mv/m’)² = Δmc²

⇒ (1/2)mv²(1+m/m’) = Δmc²

⇒

⇒

All the quantities on the RHS of the above equation are constant. Hence, v² is also constant. It shows that the emitted electron must have a constant energy. Hence, we cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus.