Class 9 Science Chapter 10 Work And Energy

Key Concept: Definition of Power

b) The speed of work done

[Solution Description]

Power is defined as the rate of doing work or the rate of transfer of energy. It can be calculated using the formula $P = \frac{W}{t}$, where $P$ is power, $W$ is work done, and $t$ is time taken.

Key Concept: Mass Calculation

b) 3 kg

[Solution Description]

The gravitational potential energy ($E_p$) is given by:

$E_p = mgh$

Rearranging the formula to solve for mass ($m$):

$m = \frac{E_p}{gh}$

Here, $E_p = 60 \, \text{J}$, $h = 2 \, \text{m}$, and $g = 10 \, \text{m s}^{-2}$. Substituting these values:

$m = \frac{60 \, \text{J}}{10 \, \text{m s}^{-2} \times 2 \, \text{m}} = 3 \, \text{kg}$

The mass of the object is $3 \, \text{kg}$.

Key Concept: Potential Energy Calculation

b) 392 J

[Solution Description]

The potential energy of an object at a height is given by the formula $E_P = mgh$. Here, $m = 5 \, \text{kg}$, $h = 8 \, \text{m}$, and $g = 9.8 \, \text{m s}^{-2}$. Substituting these values into the formula, we get:

$E_P = 5 \times 9.8 \times 8 = 392 \, \text{J}$

Therefore, the potential energy of the object is 392 J.

Key Concept: Power Definition

c) The rate of doing work

[Solution Description]

Power is defined as the rate of doing work or the rate of transfer of energy. It can be calculated using the formula $P = \frac{W}{t}$, where $W$ is the work done and $t$ is the time taken.

Key Concept: Potential Energy and Work

c) From the work done while winding it up

[Solution Description]

When you wind up a toy car, you do work on it. This work is transferred to the spring inside the car, which stores it as potential energy. The potential energy is then converted into kinetic energy when the car moves. Thus, the energy acquired by the toy car comes from the work done while winding it up.

Key Concept: Work, Energy, Force

c) 980 J

[Solution Description]

The work done against gravity is calculated using the formula $W = m \cdot g \cdot h$, where $m$ is the mass, $g$ is the gravitational acceleration, and $h$ is the height. Here, $m = 20$ kg, $g = 9.8 \, m/s^2$, and $h = 5$ m. Substituting these values, we get $W = 20 \cdot 9.8 \cdot 5 = 980$ J.

Key Concept: Positive and Negative Work

b) -48 J

[Solution Description]

When the displacement is in the opposite direction of the force, the work done is negative. The formula for work done is:

$W = F \times (-s)$

Here, $F = 12 N$ and $s = 4 m$. Substituting these values into the formula:

$W = 12 N \times (-4 m) = -48 J$

Therefore, the work done is $-48 J$.

Key Concept: Gravitational Potential Energy

b) $9.8 \, \text{m s}^{-2}$.

[Solution Description]

The gravitational potential energy ($E_p$) of an object at a height $h$ is given by the formula:

$E_p = mgh$

We are given $E_p = 784 \, \text{J}$, $m = 20 \, \text{kg}$, and $h = 4 \, \text{m}$. Rearranging the formula to solve for $g$:

$g = \frac{E_p}{mh} = \frac{784 \, \text{J}}{20 \, \text{kg} \times 4 \, \text{m}} = 9.8 \, \text{m s}^{-2}$

Therefore, the value of $g$ is $9.8 \, \text{m s}^{-2}$.

Key Concept: Energy Transformation

b) Potential energy to kinetic energy

[Solution Description]

During the free fall of an object, gravitational potential energy is converted into kinetic energy. As the object falls, its height decreases, reducing its potential energy, while its velocity increases, increasing its kinetic energy. The total mechanical energy, however, remains constant.

Key Concept: Potential Energy of an Object at a Height

c) 400 J

[Solution Description]

To calculate the potential energy of the object, we use the formula $E_p = mgh$. Here, the mass $m$ is 5 kg, the height $h$ is 8 m, and the acceleration due to gravity $g$ is 10 m/s$^2$.

Step 1: Substitute the given values into the formula.

$E_p = mgh = 5 \, \text{kg} \times 10 \, \text{m/s}^2 \times 8 \, \text{m}$

Step 2: Perform the multiplication.

$E_p = 5 \times 10 \times 8 = 400 \, \text{J}$

Therefore, the potential energy of the object is 400 J.

Key Concept: Law of Conservation of Energy

c) It remains the same

[Solution Description]

The law of conservation of energy states that the total energy of a system remains constant during energy transformation. Energy can only be converted from one form to another; it can neither be created nor destroyed. Therefore, the total energy before and after the transformation remains the same.

Key Concept: Potential and Kinetic Energy

c) 200 J

[Solution Description]

The potential energy at the top is converted into kinetic energy just before hitting the ground. The potential energy at the top is given by $PE = mgh$. Substituting the values: $PE = 2 \times 10 \times 10 = 200 \, \text{J}$. Since all this potential energy is converted into kinetic energy, the kinetic energy just before hitting the ground is also 200 J.

Key Concept: Power Calculation

c) 200 W

[Solution Description]

Power is calculated using the formula $P = \frac{W}{t}$. Here, the work done $W = 600$ J and the time taken $t = 3$ s. Substituting these values into the formula, we get:

$P = \frac{600 \, \text{J}}{3 \, \text{s}} = 200 \, \text{W}$

Therefore, the power of the machine is 200 W.

Key Concept: Energy Conversion in Nature

b) Photosynthesis in green plants

[Solution Description]

Green plants convert solar energy into chemical energy through the process of photosynthesis. This is how plants produce food and store energy.

Key Concept: Work, Force, Displacement

b) $60 J$

[Solution Description] To calculate the work done, we use the formula: $ W = F \times s $

Here, $F = 12 N$ and $s = 5 m$.

Substituting these values into the formula:

$ W = 12 N \times 5 m = 60 J $

Therefore, the work done is $60 J$.

Key Concept: Energy Sources

c) The Sun

[Solution Description]

The Sun is the biggest natural source of energy. Other sources like nuclear power plants and batteries are man-made or derived from natural sources but processed.

Key Concept: Kinetic Energy Calculation

b) 125 J

[Solution Description]

The formula for kinetic energy is given by $E_k = \frac{1}{2} m v^2$. Here, the mass ($m$) is 10 kg and the velocity ($v$) is 5 m/s. Plugging in the values, we get:

$E_k = \frac{1}{2} \times 10 \times 5^2 = \frac{1}{2} \times 10 \times 25 = 125 \, \text{J}.$

Therefore, the kinetic energy of the object is 125 J.

Key Concept: Work Done by a Constant Force

b) 10 J

[Solution Description]

Work done is calculated using the formula $W = F \times s$, where $F$ is the force applied and $s$ is the displacement in the direction of the force. Here, the force applied is equal to the weight of the book, which is 5 N, and the displacement is 2 m. Thus, the work done is $W = 5 \times 2 = 10$ J.

Key Concept: Work Done Calculation

c) 50 J

[Solution Description]

Using the formula for work done, $W = F s$, where $F = 10$ N and $s = 5$ m. Plugging in the values, we get $W = 10 \times 5 = 50$ J. Therefore, the work done is 50 joules.

Key Concept: Zero Work Done

c) When the displacement is zero

[Solution Description]

According to the definition of work, if there is no displacement of the object, the work done by the force is zero. Hence, when the displacement is zero, the work done is zero.

Key Concept: Positive and Negative Work

c) Work done by gravity is negative

[Solution Description]

When a box is lifted vertically upward, the force exerted by the person is in the direction of displacement, so the work done by the person is positive. However, gravity acts downward, opposite to the direction of displacement. Therefore, the work done by gravity is negative.

The correct statement is that the work done by gravity is negative.

Key Concept: Law of Conservation of Energy

c) $\frac{h}{4}$

[Solution Description]

According to the law of conservation of energy, the total mechanical energy remains constant during the fall. Initially, the total energy is $mgh$. At any point during the fall, the sum of kinetic energy ($KE$) and potential energy ($PE$) equals $mgh$. Given that $\frac{KE}{PE} = 3:1$, we can express $KE$ as $3PE$. The equation becomes:

$3PE + PE = mgh$

Simplifying this, we get $4PE = mgh$. Since $PE = mgh’$, where $h’$ is the height at that instant, we substitute to find:

$4mgh’ = mgh$

Cancelling out $mg$ from both sides, we get $4h’ = h$. Therefore, $h’ = \frac{h}{4}$.

Key Concept: Work and Energy

c) 100 J

[Solution Description]

Work done is calculated using the formula $W = F \times s$, where $F$ is the force and $s$ is the displacement. Here, $F = 20 \, \text{N}$ and $s = 5 \, \text{m}$. Therefore, $W = 20 \, \text{N} \times 5 \, \text{m} = 100 \, \text{J}$.

Key Concept: Power Calculation

c) 30 W

[Solution Description]

Power is calculated using the formula $P = \frac{W}{t}$, where $W$ is work done (energy consumed) and $t$ is time taken. Here, $W = 3600 \, \text{J}$ and $t = 120 \, \text{s}$. Substituting these values into the formula, we get:

$P = \frac{3600 \, \text{J}}{120 \, \text{s}} = 30 \, \text{W}$

Therefore, the power of the heater is 30 W.

Key Concept: Scientific Definition of Work

c) When a force acts on the object and the object is displaced in the direction of the force.

[Solution Description]

According to the scientific definition, work is done on an object when two conditions are satisfied: (i) a force acts on the object, and (ii) the object is displaced in the direction of the force. If either of these conditions is not met, work is not done.

Key Concept: Work Done

a) 50 J

[Solution Description]

Work done is calculated using the formula: $\text{Work} = \text{Force} \times \text{Distance}$. Here, Force = 10 N and Distance = 5 m. Therefore, $\text{Work} = 10 \text{ N} \times 5 \text{ m} = 50 \text{ J}$.

Key Concept: Law of Conservation of Energy

c) It remains constant

[Solution Description]

The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another. Therefore, the total energy remains constant during any transformation.

Key Concept: Examples of Work Not Done

c) Standing still with a heavy load on your head

[Solution Description]

According to the scientific definition of work, work is only done when a force acts on an object and the object is displaced.

In the activity where a woman stands still with a heavy load on her head, although she exerts force to hold the load, there is no displacement of the load. Therefore, no work is done.

Key Concept: Energy and Work

b) 100 J

[Solution Description]

The energy transferred to the nail is equal to the potential energy of the hammer at the height of 5 meters. The formula for potential energy is $PE = mgh$, where $m$ is mass, $g$ is acceleration due to gravity, and $h$ is height. Substituting the values, $PE = 2 \times 10 \times 5 = 100$ J. Thus, the energy transferred to the nail is 100 J.

Key Concept: Forms of Energy

c) Chemical energy

[Solution Description]

Chemical energy is the form of energy stored in the bonds between atoms and molecules. This energy is released or absorbed during chemical reactions, such as combustion or photosynthesis. The other options provided are different forms of energy that do not specifically refer to the energy stored in atomic or molecular bonds.

Key Concept: Energy Unit

c) Joule

[Solution Description]

The SI unit of energy is the joule (J). It is defined as the amount of energy required to do 1 joule of work. Other units like the calorie or kilowatt-hour are also used, but the standard SI unit is the joule. Therefore, the correct answer is joule.

Key Concept: Work Done

b) 500 J

[Solution Description]

The formula for work done is $W = Fd$. Here, $F = 50$ N and $d = 10$ m. Plugging in the values, we get:

$W = 50 \times 10 = 500 \, \text{J}$

So, the work done is 500 J.

Key Concept: Mechanical Energy, Law of Conservation of Energy

c) 200 J

[Solution Description]

When the ball is dropped, its initial potential energy is converted into kinetic energy as it falls. According to the law of conservation of energy, the total mechanical energy remains constant. The potential energy at height h is given by $PE = mgh$.

Initial potential energy: $PE_{initial} = mgh = 2 kg \times 10 m/s^2 \times 10 m = 200 J$.

At the point just before hitting the ground, all the potential energy has been converted into kinetic energy. Therefore, the kinetic energy just before impact is equal to the initial potential energy, which is 200 J.

Key Concept: Work, Force, Displacement, Angle

c) 18 J

[Solution Description]

To calculate the work done when the force is applied at an angle to the displacement, we use the formula $W = F \times s \times \cos(\theta)$, where $\theta$ is the angle between the force and displacement. Here, $F = 12$ N, $s = 3$ m, and $\theta = 60^\circ$ $\cos(60^\circ) = 0.5
$. Thus, $W = 12 \times 3 \times 0.5 = 18$ J. Therefore, the work done by the force is 18 J.