Class 11 Physics (Physics Part-I) Chapter 5 Work, Energy And Power

Key Concept: Work-Energy Theorem

a) $K_f – K_i = W $

[Solution Description] The work-energy theorem states that the work done is equal to the change in kinetic energy, which can be expressed as  $K_f – K_i = W $, where  $K_f $ and  $K_i $ are the final and initial kinetic energies respectively.

Key Concept: Conceptual Understanding

b) Sound and heat

[Solution Description] In an inelastic collision, a part of the kinetic energy is transformed into other forms such as heat and sound due to the non-conservation of kinetic energy.

Key Concept: Energy at Heights

a) $mgH$

[Solution Description]

The potential energy ($E_H$) at height $H$ for an object with mass $m$ is calculated using the formula:

$E_H = mgH$

where $g$ is the acceleration due to gravity. Thus, the potential energy is $mgH$.

Key Concept: Application of Formula

b) 150 J

[Solution Description] The work done can be calculated using the formula $W = F \cdot d$. Here, the force $F = 50 \, \text{N}$ and the displacement $d = 3 \, \text{m}$. Therefore,
$W = 50 \, \text{N} \times 3 \, \text{m} = 150 \, \text{J}.$
The work done on the box is 150 J.

Key Concept: Basic Definition

b) Work done in a closed path is zero

[Solution Description] A conservative force can be derived from a scalar potential energy function and has the property that the work done over any closed path is zero.

Key Concept: Kinetic Energy Definition

c) 6 J

[Solution Description] The formula for kinetic energy, $K$, is:
$K = \frac{1}{2}mv^2$
Substitute mass $m = 3\, \text{kg}$ and velocity $v = 2\, \text{m/s}$ into the equation:
$K = \frac{1}{2} \times 3\, \text{kg} \times (2\, \text{m/s})^2 = \frac{1}{2} \times 3 \times 4 = 6 \, \text{J}$

Key Concept: Real-World Application

d) 6 kWh

[Solution Description]
The energy consumed can be calculated using the formula: Energy (kWh) = Power (kW) $\times$ Time (h). First, convert 1500 W to kW:
$\text{Power in kW} = \frac{1500}{1000} = 1.5 \, \text{kW}$
Then calculate the energy consumed:
$\text{Energy} = 1.5 \times 4 = 6 \, \text{kWh}$
Thus, the energy consumed is 6 kWh.

Key Concept: One-Dimensional Collision

d) $\theta_1 = \theta_2 = 0$

[Solution Description] A collision is regarded as one-dimensional if both angles $\theta_1$ and $\theta_2$ are zero, meaning the motion occurs along a single line.

Key Concept: Work Done Scenarios

c) 433 J

[Solution Description] The work done is calculated as  $W = F \cdot d \cdot \cos \theta $, where  $F = 50 $ N,  $d = 10 $ m,  $\theta = 30^\circ $.

$W = 50 \times 10 \times \cos(30^\circ) = 500 \times \frac{\sqrt{3}}{2} = 250\sqrt{3} \approx 433 \text{ J} $

Key Concept: Kinetic Energy Change

b) 6.32 m/s

[Solution Description]
According to the work-energy theorem, the change in kinetic energy $\Delta K$ is equal to the work done $W$.
Initially, the kinetic energy $K_i$ is zero since the object is at rest.
So, $\Delta K = K_f – K_i = K_f = W = 100 \, \text{J}$
The formula for kinetic energy is $K_f = \frac{1}{2}mv^2$. Solving for $v$:
$100 = \frac{1}{2}(5)v^2$
Thus,
$200 = 5v^2 \Rightarrow v^2 = 40 \Rightarrow v = \sqrt{40} = 6.32 \, \text{m/s}$
Therefore, the final velocity is approximately 6.32 m/s.

Key Concept: Units and Dimensions

b) Joule (J)

[Solution Description] The SI unit of work done is the Joule (J), which is defined as one Newton meter ($1 \, N \cdot m$). It measures the amount of energy transferred or converted when applying a force over a displacement.

Key Concept: One-Dimensional Collision

c) $1.33 \, \text{m/s}$

[Solution Description]
In a completely inelastic collision, the masses stick together after the collision, and we can use the conservation of momentum to find the final velocity:
Using the formula:
$v_f = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}$
where $m_1 = 3 \, \text{kg}$, $v_{1i} = 4 \, \text{m/s}$; $m_2 = 6 \, \text{kg}$, and $v_{2i} = 0 \, \text{m/s}$,
$v_f = \frac{(3 \, \text{kg})(4 \, \text{m/s}) + (6 \, \text{kg})(0 \, \text{m/s})}{3 \, \text{kg} + 6 \, \text{kg}}$
$v_f = \frac{12 \, \text{kg}\cdot\text{m/s}}{9 \, \text{kg}}$
$v_f = 1.33 \, \text{m/s}$
Hence, the final velocity is $1.33 \, \text{m/s}$.

Key Concept: Definition Recall

b) Stored energy due to position

[Solution Description] Potential energy is the stored energy of an object due to its position or configuration.

Key Concept: Basic Definition

a) Always zero

[Solution Description]

For a conservative force, the work done along a closed path is always zero because it depends only on the initial and final positions, which are the same in a closed path.

Key Concept: Momentum Conservation

c) 20 kg·m/s

[Solution Description]

In all collisions, including elastic ones, total linear momentum is conserved. Hence, if the initial momentum is 20 kg·m/s, the final momentum will also be 20 kg·m/s.

Key Concept: Complex Problem Solving, Real-World Application

c) 240 kWh

[Solution Description] The power $P$ delivered by the engine can be calculated using the formula $P = F \cdot v$, where $F$ is the force and $v$ is the velocity. Thus, $P = 4000 \, \text{N} \times 20 \, \text{m/s} = 80000 \, \text{W}$. Now, find the total energy consumed: Energy $E = P \times t$, where $t = 3 \, \text{hours} = 3 \times 3600 \, \text{s}$; therefore, $E = 80000 \, \text{W} \times 10800 \, \text{s} = 864000000 \, \text{J}$. Converting this to kilowatt-hours: $\frac{864000000 \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} = 240 \, \text{kWh}$.

Key Concept: Basic Definition

c) The work done by the net force equals the change in kinetic energy.

[Solution Description] The work-energy theorem states that the change in kinetic energy of a body is equal to the work done by the net force on the body. This can be represented as $K_f – K_i = W_{net}$.

Key Concept: Kinetic Energy Calculation

b) 200,000 J

[Solution Description] The formula for kinetic energy ($KE$) is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. Substituting the values, we have:
$KE = \frac{1}{2} \times 1000 \, \text{kg} \times (20 \, \text{m/s})^2 = \frac{1}{2} \times 1000 \times 400 = 200000 \, \text{J}$
Therefore, the kinetic energy of the car is 200,000 J.

Key Concept: Hooke’s Law Basics

b) $F_s = -kx$

[Solution Description] According to Hooke’s Law, the force exerted by a spring is proportional to the displacement from its equilibrium position and can be expressed as $F_s = -kx$, where $k$ is the spring constant and $x$ is the displacement.

Key Concept: Gravitational Potential Energy

b) $V(h) = mgh$
[Solution Description] The formula for gravitational potential energy is given by $V(h) = mgh$, where $m$ is mass, $g$ is gravitational acceleration, and $h$ is height.

Key Concept: Positive and Negative Work

b) Negative

[Solution Description] Since the angle is $150^\circ$, and $\cos 150^\circ$ is negative, the work done will be negative because $W = F \cdot d \cdot \cos 150^\circ$ results in a negative value.

Key Concept: Hooke’s Law Understanding

b) -100 N

[Solution Description] According to Hooke’s Law, the force exerted by a spring is given by:
$F_s = -kx$
Substituting the values:
$F_s = -(250 \, \text{N/m})(0.4 \, \text{m}) = -100 \, \text{N}$
The negative sign indicates that the force is in the opposite direction to the compression.

Key Concept: Basic Calculation

c) 150 J

[Solution Description]

The work done by friction can be calculated using the formula $W = f \cdot d$. Given the frictional force $f = 30 \, \text{N}$ and displacement $d = 5 \, \text{m}$, we have:

$W = 30 \times 5 = 150 \, \text{J}$

Therefore, the work done by the frictional force is 150 joules.

Key Concept: Kinetic Energy Calculation

c) 25 Joules

[Solution Description]
The kinetic energy $K$ can be calculated using the formula:
$K = \frac{1}{2}mv^2$
Substituting the given values:
$K = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 25$
$K = 25 \, \text{Joules}$
Hence, the kinetic energy of the ball is 25 Joules.

Key Concept: Unit Vectors

b) 0

[Solution Description] The dot product of perpendicular unit vectors such as $i$ and $k$ is zero. This is because $\cos 90^\circ = 0$.

Key Concept: Zero Work Conditions

b) A person holding a heavy box stationary

[Solution Description]

According to the conditions for work to be zero, when there is no displacement or the force and displacement are perpendicular, no work is done. A person holding a heavy box stationary involves no displacement, hence no work is done.

Key Concept: Work Formula Application

d) 0 J

[Solution Description] Using the formula $W = F \cdot d \cdot \cos \theta$ where $\theta = 90^\circ$, we have:
$W = 20 \cdot 4 \cdot \cos 90^\circ = 20 \cdot 4 \cdot 0 = 0$

Key Concept: Units of Work

a) Joule

[Solution Description] The SI unit of work is the Joule, which is defined as one Newton meter. It represents the amount of work done when a force of one Newton displaces an object by one meter in the direction of the force.

Key Concept: Two-Dimensional Collision

b) $1.09 \, \text{m/s}$

[Solution Description]

In the horizontal (x) direction, momentum is conserved. The initial momentum in the x-direction is $p_{ix} = m_1 v_{1i}$. The final momentum in the x-direction for both objects is given by:

$p_{fx} = m_1 v_{1f} \cos{\theta_1} + m_2 v_{2f}$

Given that object 2 was initially at rest, we have:

$m_1 v_{1i} = m_1 v_{1f} \cos{\theta_1} + m_2 v_{2f}$

Substituting the values, $m_1 = 2\, \text{kg}$, $v_{1i} = 3\, \text{m/s}$, $v_{1f} = 2.5\, \text{m/s}$, $\theta_1 = 30^\circ$, and solving for $v_{2f}$:

$2 \times 3 = 2 \times 2.5 \times \cos{30^\circ} + m_2 v_{2f}$

$6 = 5 \times \frac{\sqrt{3}}{2} + m_2 v_{2f}$

Assume $m_2 = 2\, \text{kg}$ for calculation simplicity, then solve for $v_{2f}$:

$v_{2f} = \frac{6 – 5 \times \frac{\sqrt{3}}{2}}{2}$

$\approx 1.09 \, \text{m/s}$

Hence, the approximate value of $v_{2f}$ is $1.09 \, \text{m/s}$.

Key Concept: Complex Problem Solving, Theoretical Application

c) 5.4 s

[Solution Description]
To find the time $t$ required to reach a speed $v = 30 \, \text{m/s}$ with a constant power output $P = 100 \, \text{kW}$, we use the relationship between power and kinetic energy.
The work done by the engine is equal to the change in kinetic energy of the car:
$W = \frac{1}{2} mv^2 = \frac{1}{2} \times 1200 \, \text{kg} \times (30 \, \text{m/s})^2 = 540000 \, \text{J}$
Since power is the rate of doing work, we have:
$P = \frac{W}{t} \implies t = \frac{W}{P} = \frac{540000 \, \text{J}}{100000 \, \text{W}} = 5.4 \, \text{s}$
So, the time taken is 5.4 seconds.

Key Concept: Basic Concept

c) $W = F \cdot d \cdot \cos(\theta)$
[Solution Description]
The formula for calculating work done by a constant force is $W = F \cdot d \cdot \cos(\theta)$, where $F$ is the magnitude of the force, $d$ is the displacement, and $\theta$ is the angle between the force and the direction of displacement.

Key Concept: Basic Concept

d) $W = \int_{x_i}^{x_f} F(x) \, dx$

[Solution Description] The work done by a variable force is calculated using the definite integral of the force function over the displacement. This is expressed as $W = \int_{x_i}^{x_f} F(x) \, dx$.

Key Concept: Potential Energy Expression

b) $V(x) = \frac{kx^2}{2}$

[Solution Description] According to the formula for elastic potential energy, the potential energy $V(x)$ stored in a spring with displacement $x$ is given by $V(x) = \frac{kx^2}{2}$ where $k$ is the spring constant.

Key Concept: Unit Conversion

b) 2238 W

[Solution Description] We know that 1 horsepower is equivalent to 746 watts. To convert 3 horsepower to watts, we multiply:

$3 \text{ hp} \times 746 \text{ W/hp} = 2238 \text{ W}$

Therefore, 3 horsepower is equivalent to 2238 watts.

Key Concept: Perpendicular Vectors

a) A.B = 0

[Solution Description] When vectors A and B are perpendicular, the angle $\theta = 90^\circ$. Thus, $\cos(90^\circ) = 0$. Therefore, $A.B = AB \cos(90^\circ) = 0$.

Key Concept: Conservation Principle

c) Both x and y components of linear momentum

[Solution Description]

During any collision in a closed system, the total linear momentum is conserved. For a two-dimensional collision, this means that both the x and y components of the linear momentum are conserved.

Key Concept: Distributive Law

b) $A.(B + C) = A.B + A.C$
[Solution Description] According to the distributive law of the dot product, for vectors $A$, $B$, and $C$:
$A.(B + C) = A.B + A.C$
This means that the dot product of vector $A$ with the sum of vectors $B$ and $C$ is equal to the sum of the dot products of $A$ with each of the vectors $B$ and $C$ separately.

Key Concept: Variable Force Calculation

d) 80 J

[Solution Description] To find the work done, we integrate the force function with respect to displacement:

$W = \int_{1}^{3} 4x^3 \, dx $

Calculating the integral:

$W = \left[ x^4 \right]_{1}^{3} $

$= (3^4 – 1^4) $

$= (81 – 1) $

$= 80\, \text{J} $

Key Concept: Scalar Product Basics

b) Follows commutative law

[Solution Description] The scalar product, also known as dot product, follows the commutative law such that $A.B = B.A$. This means the order of multiplication does not affect the result.

Key Concept: Elastic vs Inelastic

a) Total kinetic energy is conserved

[Solution Description] In an elastic collision, both the total kinetic energy and the total linear momentum of the system are conserved.

Key Concept: Energy Transfer

c) $v$

[Solution Description]
According to the concept of equal mass collisions, when two equal masses collide and one comes to rest, the other takes on its initial velocity.
Given:
– Initial velocity of the first mass $v_{1i} = v$
– Final velocity of the first mass $v_{1f} = 0$
Applying conservation of momentum:
$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$
Since $m_1 = m_2$ and $v_{1f} = 0$, it simplifies to:
$v = v_{2f}$
Thus, the final velocity of the second mass is $v$.

Key Concept: Elastic vs Inelastic

a) Elastic

In elastic collisions, both linear momentum and kinetic energy are conserved due to no loss of energy as heat or sound.

Key Concept: Energy Loss

b) 168.75 J

[Solution Description]

The initial kinetic energy (\$K_i\$) of the bullet is calculated using the formula:

$K_i = \frac{1}{2} \cdot 0.005\, \text{kg} \cdot (300)^2 $

$K_i = 0.0025 \cdot 90000 = 225 \, \text{J} $

The final kinetic energy (\$K_f\$) of the bullet after emerging from the target is:

$K_f = \frac{1}{2} \cdot 0.005\, \text{kg} \cdot (150)^2 $

$K_f = 0.0025 \cdot 22500 = 56.25 \, \text{J} $

Therefore, the kinetic energy lost by the bullet is:

$K_{\text{lost}} = K_i – K_f = 225 – 56.25 = 168.75 \, \text{J} $

Key Concept: Units of Work

b) Joule

[Solution Description] The SI unit of work is ‘joule’, represented as ‘J’, and it shares the same dimensions as energy.

Key Concept: Work and Force Angle

c) 90 degrees

[Solution Description] According to the formula $W = (F \cos \theta) d$, work is zero when the angle $\theta$ between the force and displacement is 90 degrees (force and displacement are perpendicular).

Key Concept: Hooke’s Law Basics

b) $F_s = -kx$
[Solution Description]
The mathematical expression for Hooke’s Law relates the force exerted by a spring $F_s$ to its displacement from equilibrium $x$. It is given by the formula:
$F_s = -kx$
Here, $k$ is the spring constant and $x$ is the displacement.

Key Concept: Instantaneous Power

c) 150 W

[Solution Description] The formula for instantaneous power is $P = F \cdot v$. Here, $F = 30 \, \text{N}$ and $v = 5 \, \text{m/s}$. So the instantaneous power is:

$P = 30 \times 5 = 150 \, \text{W}$

Key Concept: Completely Inelastic Collision

b) They move together as one mass

[Solution Description] In a completely inelastic collision, the colliding bodies stick together and move as a single mass post-collision, resulting in maximum kinetic energy loss.

Key Concept: Formula Application

b) $v_{1f} = \frac{m_1 – m_2}{m_1 + m_2} v_{1i}$

[Solution Description] In a one-dimensional elastic collision, the final velocity $v_{1f}$ for mass $m_1$ can be calculated using the formula:
$v_{1f} = \frac{m_1 – m_2}{m_1 + m_2} v_{1i}$
where $m_1$ and $m_2$ are the masses of the two colliding objects respectively, and $v_{1i}$ is the initial velocity of mass $m_1$. This formula accounts for the conservation of both momentum and kinetic energy in elastic collisions.

Key Concept: Definition Recall

c) Stored energy due to position or configuration

[Solution Description] Potential energy is defined as stored energy due to an object’s position or configuration.

Key Concept: Zero Work Condition

c) When the force is applied perpendicular to the displacement.

[Solution Description] No work is done when the force applied is perpendicular to the displacement, since work depends on the component of force in the direction of displacement. Mathematically, $W = Fd \cos \theta$, where $\theta = 90^\circ$ results in $W = 0$ because $\cos 90^\circ = 0$.

Key Concept: Scalar Nature

c) It is a scalar quantity.

[Solution Description] Work done is a scalar quantity, meaning it has magnitude but no direction. It can be positive, negative, or zero depending on the relative directions of force and displacement.

Key Concept: Hooke’s Law Basics

b) $F_s = -kx$
[Solution Description] According to Hooke’s Law, the force exerted by a spring is directly proportional to the displacement and acts in the opposite direction of the displacement. The mathematical expression for this is $F_s = -kx$, where $F_s$ is the spring force, $k$ is the spring constant, and $x$ is the displacement.

Key Concept: Energy Conservation

b) Potential energy decreases.

[Solution Description] According to the conservation of mechanical energy, the total mechanical energy remains constant. If kinetic energy increases, potential energy must decrease to keep the sum constant.

Key Concept: Basic Concept

c) Total mechanical energy is conserved

[Solution Description] The equation $\Delta(K + V) = 0$ implies that the total mechanical energy remains constant over time, indicating the conservation of mechanical energy.

Key Concept: Formula Application

c) 22.5 J
[Solution Description] To calculate the work done, we integrate the force function $F(x) = 3x$ over the interval from $x = 1$ to $x = 4$:
$W = \int_{1}^{4} 3x \, dx$
The integral of $3x$ with respect to $x$ is $\frac{3x^2}{2}$.
$W = \left[\frac{3x^2}{2}\right]_{1}^{4}$
Evaluate at the limits:
$W = \frac{3(4)^2}{2} – \frac{3(1)^2}{2} = \frac{48}{2} – \frac{3}{2} = 24 – 1.5 = 22.5 \text{ J}$
Therefore, the work done is 22.5 Joules.

Key Concept: Basic Concept

c) Work done

[Solution Description] The area under a force-displacement graph represents the work done by the force over that displacement. It is because, for small displacements $\Delta x$, the work done $\Delta W = F(x) \Delta x$, and summing these for all displacements gives the total work.

Key Concept: Instantaneous vs Average Power, Complex Problem Solving

b) 500 W

[Solution Description]

Instantaneous power ($P$) can be calculated using the formula:

$P = F \cdot v = 100 \, \text{N} \times 5 \, \text{m/s} = 500 \, \text{W}$

Key Concept: Conservation of Momentum

b) $m_1v_{1i} = m_1v_{1f}\cos\theta_1 + m_2v_{2f}\cos\theta_2$

[Solution Description] The conservation of linear momentum in the x-direction for two-dimensional collisions is expressed as $m_1v_{1i} = m_1v_{1f}\cos\theta_1 + m_2v_{2f}\cos\theta_2$.

Key Concept: Simple Calculations

b) 8/9

[Solution Description] Given that for deuterium, $f_1 = 1/9$, the fractional kinetic energy lost $f_2$ is calculated as $f_2 = 8/9$.

Key Concept: Angle Impact

c) 22.5 J

[Solution Description]
Work done is given by $W = F \cos \theta d$. Here, $F = 15$ N, $\theta = 60$ degrees, and $d = 3$ m.
$W = 15 \times \cos(60^\circ) \times 3$
Using $\cos(60^\circ) = 0.5$, we find:
$W = 15 \times 0.5 \times 3 = 22.5 \text{ J}$
Thus, the work done is 22.5 J.

Key Concept: Basic Definition

a) Position or configuration of a body

[Solution Description] The main factors determining gravitational potential energy are mass ($m$), gravitational acceleration ($g$), and height ($h$). It depends on the position or configuration of an object.

Key Concept: Formula Application

c) 0.5

[Solution Description]
The coefficient of restitution $e$ is calculated using the formula:
$e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$
Substituting the given values:
$e = \frac{4}{8} = 0.5$
Thus, the coefficient of restitution is 0.5.

Key Concept: Energy Transfer in Collisions

d) 100%

[Solution Description]

When two objects collide elastically and one of them is initially at rest, the maximum kinetic energy transfer occurs when both masses are equal. The formula for kinetic energy transferred can be derived from the conservation of momentum and energy equations:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$
$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$
For $m_1 = m_2$ and $v_{2i} = 0$, solving these gives that the moving object transfers all its kinetic energy to the initially stationary object.
Therefore, the correct answer is: 100%.

Key Concept: Basic Concept

c) $K_f – K_i = W$

[Solution Description] The work-energy theorem connects work done by a force to the change in kinetic energy of an object. For a variable force, this is expressed as $K_f – K_i = W$, where $K_f$ is the final kinetic energy, $K_i$ is the initial kinetic energy, and $W$ is the work done.

Key Concept: Distributive Law

a) $A.(B + C) = A.B + A.C$

[Solution Description] In the distributive law, the dot product should distribute over addition: $A.(B + C) = A.B + A.C$. This means that if you distribute vector $A$ over the sum of vectors $B$ and $C$, you take the dot product of $A$ with each separately and then sum the results.

Key Concept: Energy Conservation

c) Total mechanical energy is conserved throughout the motion.

[Solution Description] For a pendulum with only conservative forces acting, the total mechanical energy remains constant throughout the pendulum’s motion.

Key Concept: Equal Masses Scenario

c) 8 m/s

[Solution Description] For equal masses ($m_1 = m_2$), $v_{1f} = 0$ and $v_{2f} = v_{1i}$. Therefore, since $v_{1i} = 8 \, \text{m/s}$, we have:

$v_{2f} = v_{1i} = 8 \, \text{m/s}.$