work and energy class 9

Page:114

Q.1.When do we say that work is done?

Ans=Work is done when a force causes a displacement in the direction of the force applied, and the magnitude of the force and the displacement are not perpendicular to each other.

Q.2.Write an expression for the work done when a force is acting on an object in the direction of its displacement?

Ans=The work done ($$) when a force ($F$) acts on an object in the direction of its displacement ($s$) is given by the formula:

$W=F\cdot s$

where:

• $W$ is the work done,
• $F$ is the magnitude of the force applied, and
• s is the displacement of the object in the direction of the force. 

Q.3. Define 1 J of work.

Ans=One joule (1 J) of work is defined as the amount of work done when a force of 1 newton is applied to move an object through a distance of 1 meter in the direction of the force. Mathematically, 1 joule is equivalent to $1\text{J}=1\text{N}\cdot \text{m}$.

 
Q.4.A pair of bullocks exerts a force of 140 N on a plough.  The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Ans=The work done ($W$) is calculated using the formula:

$W=F\cdot$s

where:

• $F$ is the force applied,
• s is the displacement.

In this case, $F=140\text{N}$(force exerted by the pair of bullocks) and $s=15\text{m}$ (length of the field being ploughed).

$W=140\text{N}\cdot 15\text{m}$

$W=2100\text{N}\cdot \text{m}$

Therefore, the work done in ploughing the length of the field is $2100\text{N}\cdot \text{m}$, which is equivalent to $2100\text{J}$(joules).

page119

Q.1.What is the kinetic energy of an object?

Ans= The kinetic energy ($E_k$) of an object is the energy it possesses due to its motion. 

Q.2.Write an expression for the kinetic energy of an object?

Ans=

The formula for kinetic energy is given by:

$E_k=\frac{1}{2}m{v}^2$

where:

• $E_k$ is the kinetic energy,
• $m$ is the mass of the object,
• $v$ is the velocity of the object.

In words, kinetic energy is proportional to the mass of the object and the square of its velocity. This means that an object with more mass or higher velocity will have a greater kinetic energy.

Q.3.The kinetic energy of an object of mass, m moving with a velocity of 5 m s–1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Ans=

The formula for kinetic energy ($E_k$) is given by:

$E_k=\frac{1}{2}m{v}^2$

where:

• $m$ is the mass of the object,
• $v$ is the velocity of the object.

Given that the initial kinetic energy ($E_{k1}$) is 25 J and the initial velocity ($v_1$) is 5 m/s, we can use this information to find the mass ($m$):

$25=\frac{1}{2}m(5{)}^2$

Solving for $m$:

$25=\frac{1}{2}\times 25m$

$50=25m$

$m=2\text{kg}$

Now, we can find the new kinetic energy ($E_{k2}$) when the velocity is doubled ($v_2=2\times v_1$):

$E_{k2}=\frac{1}{2}\times 2\text{kg}\times (2\times 5\text{m/s}{)}^2$

$E_{k2}=\frac{1}{2}\times 2\text{kg}\times 100{\text{m}}^2\mathrm{/}{\text{s}}^2$

$E_{k2}=100\text{J}$

Therefore, when the velocity is doubled, the kinetic energy becomes 100 J.

Similarly, when the velocity is increased three times ($v_3=3\times v_1$):

$E_{k3}=\frac{1}{2}\times 2\text{kg}\times (3\times 5\text{m/s}{)}^2$

$E_{k3}=\frac{1}{2}\times 2\text{kg}\times 225{\text{m}}^2\mathrm{/}{\text{s}}^2$

$E_{k3}=225\text{J}$

$\text{<b>page 123</b>}$

$\text{Q.1.What is power<b>?</b>}$

$\text{<b>Ans= </b>}$$\text{<br>}$

Power is the rate at which work is done or the rate at which energy is transferred or converted. Mathematically, power ($P$) is defined as the amount of work done ($W$) or energy transferred ($E$) per unit of time ($t$). The formula for power is:

$P=\frac{W}{t}$

or

$P=\frac{E}{t}$

where:

• $P$ is power,
• $W$ is work done,
• $E$ is energy,
• $t$ is time.

The unit of power in the International System of Units (SI) is the watt (W), where 1 watt is equal to 1 joule per second.

 Q.2.Define 1 watt of power.

Ans=One watt (1 W) of power is defined as the rate at which work is done or energy is transferred or converted when one joule of work is done in one second. In simple terms, 1 watt is equivalent to the energy transfer of 1 joule per second. 

Q.3.A lamp consumes 1000 J of electrical energy in 10 s.  What is its power? 

Ans=

The power ($P$) of the lamp can be calculated using the formula:

$P=\frac{E}{t}$

where:

• $P$ is power,
• $E$ is energy consumed,
• $t$ is time.

In this case, the lamp consumes $E=1000\text{J}$ of electrical energy in $t=10\text{s}$. Plugging these values into the formula:

$P=\frac{1000\text{J}}{10\text{s}}$

$P=100\text{W}$

Therefore, the power of the lamp is $100\text{W.}$

$\text{<b>Q.4. Define average power. </b>}$

$\text{<b>Ans=</b>}$Average power is the amount of power averaged over a given period of time. It is calculated by dividing the total energy consumed or the total work done by the total time taken. 

 

excrise 

Q.1.Look at the activities listed below.  Reason out whether or not work is done in the light of your understanding of the term ‘work’.
•Suma is swimming in a pond.
•A donkey is carrying a load on its back.
•A wind-mill is lifting water from a well.
•A green plant is carrying out photosynthesis.
•An engine is pulling a train.
•Food grains are getting dried in the sun.
•A sailboat is moving due to wind energy.

$\text{ Ans=}$

• Suma is swimming in a pond:

  • Work is done. If Suma changes her place during swimming, there is a displacement, and work is done.

• A donkey is carrying a load on its back:

  • No work is done.
    

• A windmill is lifting water from a well:

  • Work is done. The windmill exerts a force to lift the water against gravity, causing a displacement in the upward direction.

• A green plant is carrying out photosynthesis:

  • Photosynthesis involves the conversion of energy but doesn't necessarily involve work in the scientific sense. Work, in physics, typically involves a force causing a displacement. While energy transformations occur during photosynthesis, it may not fit the strict definition of work.

• An engine is pulling a train:

  • Work is done. The force exerted by the engine causes the train to move in the direction of the force applied.

• Food grains are getting dried in the sun:

  • No work is done on the food grains by the sun. In the scientific sense, work involves a force causing a displacement, and the sun is not exerting a force on the food grains to move them.

• A sailboat is moving due to wind energy:

  Work is done. The wind applies a force to the sail, causing the boat to move. The force and the displacement are in the same direction. 

2.An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

 

Ans=The work done by the force of gravity on the object is zero.

In a situation where the initial and final points of the object's path lie on the same horizontal line, the displacement of the object is horizontal. Work is defined as the product of force and displacement in the direction of the force. The force of gravity acts vertically downward, but the displacement is horizontal. Since the force and displacement are perpendicular in this case, the work done by gravity is zero.

Mathematically, the work ($W$) done by a force ($F$) over a displacement ($d$) is given by:

$W=F\cdot d\cdot \cos (\theta )$

where $\theta$ is the angle between the force and the displacement vectors. If $\theta$ is 90 degrees (perpendicular), the cosine of 90 degrees is zero, making the work done zero.

3.A battery lights a bulb. Describe the energy changes involved in the process.

Ans=

• Chemical Energy: The battery stores chemical energy.

• Electrical Energy: The battery converts chemical energy into electrical energy.

• Radiant and Thermal Energy: The bulb transforms electrical energy into light (radiant energy) and some heat (thermal energy).

In simple terms, the battery provides the energy, and the bulb transforms it into light and a little heat.

4.Certain force acting on a 20 kg mass changes its velocity from 5 m s–1  to 2 m s–1.  Calculate the work done by the force?

Answer = Given:

• Initial velocity ($u$) = 5 m/s
• Mass ($m$) = 20 kg
• Final velocity ($v$) = 2 m/s

The initial kinetic energy ($E_{\text{initial}}$) is calculated as: $E_{\text{initial}}=\frac{1}{2}\times m\times u^2=\frac{1}{2}\times 20\text{kg}\times (5\text{m/s}{)}^2=250$J

 

The final kinetic energy ($E_{\text{final}}$) is calculated as:

$E_{\text{final}}=\frac{1}{2}\times m\times v^2=\frac{1}{2}\times 20\text{kg}\times (2\text{m/s}{)}^2=40\text{J}$

$$

The work done ($W$), which is the change in kinetic energy, is given by: $W=E_{\text{final}}-E_{\text{initial}}=40\text{J}-250\text{J}=-210\text{J}$

 

5. A mass of 10 kg is at a point A on a table.  It is moved to a point B.  If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer. 

Ans=

Given:

• Mass ($m$) = 10 kg
• Acceleration due to gravity ($g$) ≈ 9.8 m/s²
• Vertical height ($h$) = 0 (horizontal displacement)

The formula for work done by the gravitational force ($W$) is:

$W=mgh$

Substitute the given values:

$W=(10\text{kg})\cdot (9.8{\text{m/s}}^2)\cdot (0\text{m})$

$W=0$

$$Since the vertical height ($h$) is zero, the work done ($W$) is zero. Therefore, the gravitational force does no work on the object when it is moved horizontally from point A to point B.

(The formula for work involves the displacement of the object and the force applied in the direction of the displacement. In this situation, the force of gravity acts vertically downward, but the displacement is horizontal. Therefore, the angle between the force of gravity and the displacement is 90 degrees, and the cosine of 90 degrees is zero.)

 6. The potential energy of a freely falling object decreases progressively.  Does this violate the law of conservation of energy? Why? 

Ans=

No, the decrease in potential energy of a freely falling object does not violate the law of conservation of energy. The law of conservation of energy states that the total energy of an isolated system remains constant over time. Energy can change from one form to another, but the total amount of energy remains constant.

In the case of a freely falling object, as it descends under the influence of gravity, its potential energy decreases. However, this decrease in potential energy is accompanied by an increase in kinetic energy. As the object falls, it gains speed, and kinetic energy is converted from potential energy.

The sum of kinetic energy and potential energy of the object at any given point in time remains constant, neglecting any energy losses due to factors like air resistance. This conservation of mechanical energy is an expression of the law of conservation of energy.

Mathematically, for an object in a gravitational field near the Earth's surface, the sum of potential energy ($U$) and kinetic energy ($K$) is constant:

$U+K=\text{constant}$

So, the decrease in potential energy is compensated by the increase in kinetic energy, and the total mechanical energy of the system is conserved. The law of conservation of energy is not violated in this process.

 7.What are the various energy transformations that occur when you are riding a bicycle?

Ans=

When riding a bicycle, various energy transformations take place as the body and the bicycle interact with the environment. Here are the key energy transformations involved:

Muscular Energy to Mechanical Energy (Pedaling):

• The primary source of energy for riding a bicycle comes from the rider's muscles. Muscular energy is converted into mechanical energy as the rider pedals, turning the pedals and driving the bicycle chain.

Mechanical Energy to Kinetic Energy (Forward Motion):

• As the rider pedals and the bicycle moves forward, the mechanical energy is converted into kinetic energy of the bicycle and the rider. This is evident in the forward motion of the entire system.

Potential Energy to Kinetic Energy (Descending):

• When the cyclist is riding downhill or descending, potential energy is converted into kinetic energy. As the bicycle moves to a lower altitude, it gains speed, and the potential energy due to height is transformed into kinetic energy.

Kinetic Energy to Mechanical Energy (Braking):

• When the brakes are applied, the kinetic energy of the moving bicycle is converted back into mechanical energy. This energy is then dissipated as heat due to the friction between the brake pads and the wheel rims.

Kinetic Energy to Sound and Heat (Friction):

• Some of the kinetic energy is transformed into sound and heat as a result of friction between the bicycle tires and the road surface. This is especially noticeable when braking or when there is resistance to motion.

In summary, riding a bicycle involves a series of energy transformations, including the conversion of muscular energy to mechanical energy, potential energy to kinetic energy, kinetic energy to mechanical energy, and the various conversions associated with friction and braking. These transformations demonstrate the principles of energy conservation and the different forms of energy at play during a bicycle ride.

8.Does the transfer of energy take place when you push a huge rock with all your might and fail to move it?  Where is the energy you spend going?

Ans=

When you push a huge rock with all your might and fail to move it, there is indeed a transfer of energy, but the energy is not doing the work of moving the rock over a distance. Instead, the energy you spend is primarily transformed into other forms.

The energy you apply while pushing the rock is converted into several different forms:

Mechanical Work Against Friction:

• A significant portion of the energy you exert is likely converted into overcoming static friction between the rock and the surface on which it rests. This results in heating of the contact surfaces and dissipation of energy as thermal energy.

Deformation Energy:

• If there is any deformation in the rock or the surface, energy may be spent in deforming the materials. This could include compressing the ground, deforming the shape of the rock, or causing other changes in the materials involved.

Sound Energy:

• Some of the energy may be transformed into sound as the pushing action creates vibrations and disturbances in the surrounding air and materials.

Elastic Potential Energy:

• If there is any elasticity in the materials involved, such as the compression of the ground or deformation of the rock, some of the energy may be stored temporarily as elastic potential energy.

In summary, when you push a huge rock and fail to move it, the energy you expend is not used to perform the work of displacing the rock. Instead, it is transformed into various other forms, such as overcoming friction, causing deformation, producing sound, and storing elastic potential energy. The inability to move the rock indicates that the energy you applied was not sufficient to overcome the resistive forces acting on the rock.

 

 9.A certain household has consumed 250 units of energy during a month.  How much energy is this in joules?

Ans=

To convert energy consumption from units to joules, you need to know the conversion factor. The most common unit for energy in the context of household electricity is the kilowatt-hour (kWh), where 1 kWh is equal to 3.6 million joules.

Given that 1 kilowatt-hour (kWh) = 3.6 x 10^6 joules, you can calculate the energy consumption in joules using the formula:

Let's calculate:

$250\text{units=250kWh}$

$\; =250\times 3.6\times 10^6\text{joules}$

$= 900\times 10^6\text{joules}$

Therefore, the household has consumed 900 million joules of energy during the month.

 

10.An object of mass 40 kg is raised to a height of 5 m above the ground.  What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. 

Ans=

The potential energy ($U$) of an object in a gravitational field is given by the formula:

$U=mgh$

where:

• $m$ is the mass of the object (40 kg),
• $g$ is the acceleration due to gravity (approximately 9.8 m/s²),
• $h$ is the height above the reference point (5 m).

Let's calculate the potential energy:

$U=(40\text{kg})\times (9.8{\text{m/s}}^2)\times (5\text{m})$

$U=1960\text{Joules}$

So, the potential energy of the object when it is raised to a height of 5 m above the ground is 1960 Joules.

Now, when the object is halfway down, it has lost half of its potential energy, and this energy is converted into kinetic energy ($K$) according to the conservation of energy.

The kinetic energy is given by the formula:

$K=\frac{1}{2}m{v}^2$

At the halfway point, all of the potential energy has been converted into kinetic energy. Therefore,

$\frac{1}{2}m{v}^2=\frac{1}{2}U$

Solving for $v$ (velocity):

$v^2=\frac{U}{m}$

Substitute the values:

$v^2=\frac{1960\text{J}}{40\text{kg}}$

$v^2=49{\text{m}}^2\mathrm{/}{\text{s}}^2$

$v=7\text{m/s}$

the kinetic energy ($K$) at the halfway point using the obtained velocity ($v=7\text{m/s}$).

The kinetic energy formula is given by:

$K=\frac{1}{2}m{v}^2$

Substitute the values:

$K=\frac{1}{2}\times 40\text{kg}\times (7\text{m/s}{)}^2$

$K=\frac{1}{2}\times 40\text{kg}\times 49{\text{m}}^2\mathrm{/}{\text{s}}^2$

$K=980\text{Joules}$

Therefore, at the halfway point, the kinetic energy of the object is 980 Joules.

 

11.What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer

Ans=

The work done by the force of gravity on a satellite moving around the Earth is zero. This is because the gravitational force is acting perpendicular to the direction of the satellite's motion.

The formula for work ($W$) is given by:

$W=F\cdot d\cdot \cos (\theta )$

where:

• $F$ is the force applied,
• $d$ is the displacement,
• $\mathrm{\theta }$is the angle between the force and the displacement.

In the case of a satellite moving in a circular orbit around the Earth, the force of gravity is directed toward the center of the Earth, and the satellite's displacement is tangential to its circular path. This means that the angle ($\theta$) between the force of gravity and the displacement is 90 degrees.

Since, $\cos (90^{\circ })=0$, the term $\cos (\theta )$ becomes zero, and the work done ($W$) is zero.

Therefore, the work done by the force of gravity on a satellite moving around the Earth is zero, and this is justified by the fact that the force and the displacement are perpendicular to each other. The gravitational force does not contribute to the satellite's change in kinetic energy or the doing of work in this context.

 

12.Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

Ans= Write it according to you...

      (In classical physics, the concept of displacement is closely tied to the application of force. According to Newton's first law of motion, an object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by a net external force. This implies that a force is needed to cause a change in the state of motion of an object.

Therefore, in the absence of any force acting on an object, according to Newton's first law, the object will remain at rest or continue to move with a constant velocity. In this scenario, there would be no net displacement of the object.

However, it's essential to note that in certain situations, there may be apparent displacement or motion even in the absence of an external force. For example:

1. Inertia and Initial Conditions:

   • If an object is already in motion or at rest and no external force is applied, it will maintain its state due to inertia. This may give the appearance of displacement, but it's a continuation of the initial conditions rather than a result of a force acting at that moment.

2. Absence of Net Force:

   • In a system where forces cancel each other out, there might not be a net force acting on an object, and the object would experience no acceleration. In such a case, the object would continue with its current state of motion, and there might be no apparent displacement.

It's important to consider the context and specifics of the scenario to fully understand the relationship between force and displacement. In classical mechanics, displacement is typically associated with the application of force, but there can be cases where apparent motion or lack of displacement is observed in the absence of a net external force.)

 

13.A person holds a bundle of hay over his head for 30 minutes and gets tired.  Has he done some work or not? Justify your answer

Ans= Write it by yourself....

(In the context of physics, work is defined as the transfer of energy that occurs when a force is applied to an object and the object moves in the direction of the force. The formula for work ($W$) is given by:

$W=F\cdot d\cdot \cos (\theta )$

where:

• $W$ is the work done,
• $F$ is the force applied,
• $d$ is the displacement,
• $\theta$ is the angle between the force and the displacement.

In the scenario you described, the person is holding a bundle of hay over their head, and there is no vertical displacement of the hay. The force applied by the person is upward (against gravity), but the displacement is zero because the bundle of hay is not moving vertically.

Since the displacement ($d$) is zero, the work done ($W$) is also zero. In other words, the person is exerting a force to hold the bundle of hay in place, but there is no work done on the hay because there is no vertical displacement.

While the person may feel tired, this fatigue is associated with the exertion of force rather than the performance of work in the physics sense. Work requires both force and displacement in the direction of the force, and in this case, the hay is not being lifted or moved vertically.)

 

To be continued........ (Cheeck again later)