1. Write detailed answers?
a. Explain the difference between potential energy and kinetic energy.
Answer
Energy is the capacity of an object to do work. Potential energy and kinetic energy are two important forms of mechanical energy.
| Potential Energy | Kinetic Energy |
|---|---|
| Potential energy is the energy possessed by an object due to its position or configuration. | Kinetic energy is the energy possessed by an object due to its motion. |
| It depends on the height, position, or arrangement of the object. | It depends on the mass and velocity of the moving object. |
| A stationary object can possess potential energy. | An object must be in motion to possess kinetic energy. |
| Gravitational potential energy is given by PE = mgh. **Kinetic e | nergy is given by** KE =(1/2)mv². |
| Example: Water stored in a dam possesses gravitational potential energy. | Example: Flowing water possesses kinetic energy. |
Explanation
A stone held at a height above the ground has potential energy because of its position in Earth's gravitational field. When the stone falls, its potential energy decreases and is converted into kinetic energy.
Similarly, a moving bicycle has kinetic energy because it is in motion. If the bicycle stops, its kinetic energy becomes zero.
Thus, potential energy is associated with position or configuration, whereas kinetic energy is associated with motion.
b. Derive the formula for the kinetic energy of an object of mass m, moving with velocity v.
Answer
Let an object of mass mbe initially at rest. A constant force Facts on it and moves it through a displacement s. Its final velocity becomes v.
We need to derive the formula for its kinetic energy.
Given:
Mass of the object = m
Initial velocity, u = 0
Final velocity, v
Acceleration = a
Displacement = s
formula for work done.
W = Fs
According to Newton's second law of motion:
F = ma
Therefore,
W = mas
Use the equation of motion.
We know that:
v²=u²+ 2as
Since the object starts from rest, u = 0.
v²= 2as
Rearranging:
as =(v²/2)
Step 3: Substitute the value of asin the work equation.
W = m( (v²/2) )
W =(1/2)mv²
The work done on the object increases its kinetic energy. Hence,
[KE =(1/2)mv²]
Therefore, the kinetic energy of an object of mass m, moving with velocity v, is equal to half the product of its mass and the square of its velocity.
The SI unit of kinetic energy is the joule (J).
c. Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.
Answer
Consider an object of mass mfalling freely from a height habove the ground. Assume that air resistance is negligible.
Let:
Mass of the object = m
Initial height = h
Initial velocity, u = 0
Acceleration due to gravity = g
Final velocity just before reaching the ground = v
Calculate the initial potential energy.
The initial potential energy of the object is:
PE = mgh
Find the velocity of the object when it reaches the ground.
Using the equation of motion:
v²=u²+ 2gh
Since u = 0,
v²= 2gh
Calculate the kinetic energy at the ground.
The kinetic energy is:
KE =(1/2)mv²
Substituting v²= 2gh:
KE =(1/2)m( 2gh )
KE = mgh
Compare the energies.
KE = PEinitial
Therefore,
[Initial potential energy=Kinetic energy at the ground]
Conclusion: When an object falls freely under gravity, its initial gravitational potential energy is converted into kinetic energy. The total mechanical energy remains constant, provided air resistance and other energy losses are neglected.
d. Determine the amount of work done when an object is displaced at an angle of 300 with respect to the direction of the applied force.
Answer
Note: The angle written as 300 is interpreted as 30°.
When a force acts on an object and the object is displaced at an angle to the force, the work done is calculated using:
W = Fs cosθ
Where:
W= Work done
F= Applied force
s= Displacement
θ= Angle between the force and displacement
For θ =30°:
W = Fs cos 30°
Since,
cos 30°=(√(3)/2)
Therefore,
[W =(√(3)/2)Fs]
or approximately,
[W = 0.866Fs]
Conclusion: The work done is approximately 86.6% of the product of the applied force and displacement.
A numerical value cannot be determined unless the magnitudes of the force and displacement are given.
e. If an object has 0 momenta, does it have kinetic energy? Explain your answer.
Answer
No, an object with zero momentum has zero kinetic energy, provided its mass is non-zero.
Momentum is given by:
p = mv
If the momentum is zero:
p = 0
For an object having non-zero mass:
mv = 0
Therefore,
v = 0
The kinetic energy of an object is:
KE =(1/2)mv²
Substituting v = 0:
KE =(1/2)m( 0 )²
[KE = 0]
Conclusion: A stationary object with zero momentum has no kinetic energy because its velocity is zero.
f. Why is the work done on an object moving with uniform circular motion zero?
Answer
In uniform circular motion, an object moves along a circular path with constant speed. Its direction of motion changes continuously.
The centripetal force acting on the object is directed towards the centre of the circle. At every point, the instantaneous displacement is tangential to the circle.
Thus, the centripetal force and instantaneous displacement are perpendicular to each other.
The formula for work done is:
W = Fs cosθ
Here,
θ =90°
Since,
cos 90°= 0
Therefore,
W = Fs( 0 )
[W = 0]
Conclusion: The centripetal force does no work during uniform circular motion because it is always perpendicular to the instantaneous displacement. The speed and kinetic energy remain constant.
2. Choose one or more correct alternatives.
a. For work to be performed, energy must be ….
(i) transferred from one place to another
(ii) concentrated
(iii) transformed from one type to another
(iv) destroyed
Answer
Correct alternatives: (i) and (iii)
Explanation: Work involves the transfer of energy or the transformation of energy from one form to another. Energy is neither created nor destroyed, according to the law of conservation of energy.
b. Joule is the unit of …
(i) force
(ii) work
(iii) power
(iv) energy
Answer
Correct alternatives: (ii) and (iv)
Explanation: The joule (J) is the SI unit of both work and energy. Force is measured in newtons (N), while power is measured in watts (W).
c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?
(i) the horizontal applied force
(ii) gravitational force
(iii) reaction force in vertical direction
(iv) force of friction
Answer
Correct alternatives: (i), (ii), (iii), and (iv), under the usual assumptions of the question.
Explanation: For an object being dragged at constant velocity on a horizontal surface:
The horizontal applied force and frictional force have equal magnitudes and opposite directions.
The gravitational force and the vertical reaction force have equal magnitudes and opposite directions.
Thus, the applied force equals friction, and the gravitational force equals the normal reaction.
d. Power is a measure of the …….
(i) the rapidity with which work is done
(ii) amount of energy required to perform the work
(iii) The slowness with which work is performed
(iv) length of time
Answer
Correct alternative: (i)
Explanation: Power is the rate at which work is done or energy is transferred.
P =(W/t)
A higher power means that the same amount of work can be performed in less time.
e. While dragging or lifting an object, negative work is done by
(i) the applied force
(ii) gravitational force
(iii) frictional force
(iv) reaction force
Answer
Correct alternatives: (ii) and (iii), depending on the motion and the forces involved.
Explanation:
Gravitational force: When an object is lifted upward, gravity acts downward, opposite to the displacement. Hence, gravity does negative work.
Frictional force: When an object slides along a surface, friction generally acts opposite to its displacement, doing negative work.
The applied force generally does positive work when it acts in the direction of displacement.
3. Rewrite the following sentences using a proper alternative.
a. The potential energy of your body is least when you are …..
(i) sitting on a chair
(ii) sitting on the ground
(iii) sleeping on the ground
(iv) standing on the ground
Answer
Correct alternative: (iii) sleeping on the ground
Rewritten sentence: The potential energy of your body is least when you are sleeping on the ground.
Explanation: Gravitational potential energy depends on height:
PE = mgh
When the body is closest to the ground, its height is least, so its gravitational potential energy is least relative to the chosen reference level.
b. The total energy of an object falling freely towards the ground …
(i) decreases
(ii) remains unchanged
(iii) increases
(iv) increases in the beginning and then decreases
Answer
Correct alternative: (ii) remains unchanged
Rewritten sentence: The total energy of an object falling freely towards the ground remains unchanged.
Explanation: During free fall, potential energy is converted into kinetic energy. The total mechanical energy remains constant if air resistance is neglected.
c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ….
(i) will be twice its original energy
(ii) will not change
(iii) will be 4 times its original energy
(iv) will be 16 times its original energy.
Answer
Correct alternative: (ii) will not change
Rewritten sentence: If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy will not change.
Explanation: The car is moving on a flat surface, so its height remains constant. Gravitational potential energy depends on height, not velocity.
However, its kinetic energy would increase 16 times because kinetic energy is proportional to the square of velocity.
d. The work done on an object does not depend on ….
(i) displacement
(ii) applied force
(iii) initial velocity of the object
(iv) the angle between force and displacement.
Answer
Correct alternative: (iii) initial velocity of the object
Rewritten sentence: The work done on an object does not depend directly on the initial velocity of the object.
Explanation: Work done by a constant force is given by:
W = Fs cosθ
It depends on the applied force, displacement, and angle between them. Initial velocity is not directly present in this formula.
4. Study the following activity and answer the questions.
Question
Questions
Answer 1. At the moment of releasing the balls, which energy do the balls have?
At the moment of release, the balls possess gravitational potential energy because they are held at a height above the floor.
Their potential energy is:
PE = mgh
Since both balls have the same mass and are released from the same height, they possess equal gravitational potential energy, assuming the same reference level.
Answer 2. As the balls roll down which energy is converted into which other form of energy?
As the balls roll down the channels, their gravitational potential energy is converted into kinetic energy.
The kinetic energy includes:
Translational kinetic energy due to the motion of the balls.
Rotational kinetic energy due to the rolling motion.
Some energy may also be converted into heat and sound because of friction and other resistive effects.
Answer 3. Why do the balls cover the same distance on rolling down?
The two balls are of the same mass and size, and they are released from the same height. Therefore, they possess equal initial gravitational potential energy.
As they roll down, their potential energy is converted into kinetic energy. If the channels have similar rolling conditions and energy losses are negligible, the balls acquire comparable speeds at the bottom.
Consequently, both balls can cover the same distance under similar conditions.
Answer 4. What is the form of the eventual total energy of the balls?
At the end of the motion, the initial gravitational potential energy has been converted mainly into kinetic energy, including translational and rotational kinetic energy.
Some energy may also have been transformed into heat and sound due to friction.
Therefore, the total energy is conserved, although its form changes.
Answer 5. Which law related to energy does the above activity demonstrate? Explain.
The activity demonstrates the law of conservation of energy.
This law states that energy can neither be created nor destroyed. It can only be transformed from one form into another.
In this activity:
Potential energy→Kinetic energy
The gravitational potential energy of the balls at the top of the channels changes into kinetic energy as they roll downward. The total energy remains conserved when all forms of energy are considered.
5. Solve the following examples.
a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m? (Ans : 1224.5 kg)
Solution
Given:
Power of the pump, P = 2 kW= 2000 W
Height, h = 10 m
Time, t = 1 minute= 60 s
Acceleration due to gravity, g = 9.8 m/s²
We need to find the mass of water lifted every minute.
Step 1: Use the formula for power.
P =(W/t)
Therefore,
W = Pt
Step 2: Calculate the work done in lifting water.
The work done against gravity is:
W = mgh
Therefore,
mgh = Pt
Rearranging:
m =(Pt/gh)
Step 3: Substitute the values.
m =(2000 × 60/9.8 × 10)
m =(120000/98)
m = 1224.49 kg
[m = 1224.5 kg]
Final answer: The pump can lift approximately 1224.5 kg of water every minute to a height of 10 m, assuming 100% efficiency.
b. If a 1200 W electric iron Is used daily for 30 minutes, how much total electricity is consumed in April? ( 18 unit)
Solution
Given:
Power of the electric iron, P = 1200 W= 1.2 kW
Daily usage = 30 minutes = 0.5 hour
Number of days in April = 30 days
We need to calculate the total electrical energy consumed.
Step 1: Calculate daily energy consumption.
E = Pt
E = 1.2 × 0.5
E = 0.6 kWh
Step 2: Calculate energy consumption for April.
Total energy= 0.6 × 30
Total energy= 18 kWh
Since,
1 unit= 1 kWh
Therefore,
[Total electricity consumed= 18 units]
Final answer: 18 units of electricity.
c. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound? (Ans : 6 m)
Solution
Given:
Initial height of the ball, h₁= 10 m
Energy lost = 40%
We need to find the rebound height.
Step 1: Calculate the remaining energy.
If 40% of the energy is lost, the remaining energy is:
100% - 40% = 60%
The initial potential energy is:
PE₁= mgh₁
The energy remaining after the collision is:
PE₂= 0.60mgh₁
Step 2: Relate the remaining energy to rebound height.
At the maximum rebound height, the ball's kinetic energy is momentarily zero, so its energy is gravitational potential energy.
mgh₂= 0.60mgh₁
Cancel mgfrom both sides:
h₂= 0.60h₁
Step 3: Substitute the initial height.
h₂= 0.60 × 10
[h₂= 6 m]
Final answer: The ball will rebound to a height of 6 m, assuming the remaining energy is entirely converted into gravitational potential energy.
d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg? (Ans. : 131250 J)
Solution
Given:
Mass of the car, m = 1500 kg
Initial velocity, u = 54 km/hr
Final velocity, v = 72 km/hr
We need to calculate the work done in increasing the car's velocity.
Convert velocities into m/s.
Since,
1 km/hr=(5/18) m/s
Initial velocity:
u = 54 ×(5/18)
u = 15 m/s
Final velocity:
v = 72 ×(5/18)
v = 20 m/s
Use the work-energy theorem.
The net work done on an object is equal to the change in its kinetic energy.
W = KEf- KEi
W =(1/2)mv²-(1/2)mu²
Taking (1/2)mcommon:
W =(1/2)m( v²-u² )
Substitute the values.
W =(1/2)× 1500 ×( 20²-15² )
W = 750 ×( 400 - 225 )
W = 750 × 175
[W = 131250 J]
Final answer: The work done on the car is 131250 J.
e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi? (Ans: 3 J)
Solution
Given:
Applied force, F = 10 N
Displacement, s = 30 cm= 0.30 m
Angle between force and displacement, θ =0°
Write the formula for work done.
W = Fs cosθ
Substitute the values.
W = 10 × 0.30 ×cos 0°
Since,
cos 0°= 1
Therefore,
W = 10 × 0.30 × 1
W = 3 J
[W = 3 J]