Class 9 Science(Exploration) Chapter 7 - Work, Energy, and Simple Machines Question Answers: Download Solution PDF

NCERT Solutions for Class 9 Science Chapter 7 Question Answers PDF is prepared for students to understand the chapter in a clear and simple way. In this NCERT solution, you get well-explained answers to all questions, covering important topics like work, energy, power, and simple machines. This PDF is useful for daily practice as well as quick revision before exams. It is a helpful resource to strengthen your concepts and score better marks in tests.

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NCERT Solution Class 9 Science Chapter 7 - Work, Energy, and Simple Machines: Revise, Reflect, Refine Question Answer

In this section, you will get the simple and clear answers to the Revise, Reflect, Refine questions of the chapter. It helps students revise important concepts and check their understanding effectively before exams.

1. State whether True or False.
(i) Work is said to be done when a force is applied, even if the object does not move.
(ii) Lifting a bucket vertically upward results in positive work done on the bucket.
(iii) The SI unit for both work and energy is joule (J).
(iv) A motionless stretched rubber band has kinetic energy.
(v) Energy can change from one form to another.

Answer:

(i) False
Work is done only when force causes displacement. If there is no movement, no work is done.

(ii) True
When a bucket is lifted upward, force and displacement are in the same direction, so work done is positive.

(iii) True
The SI unit of both work and energy is joule (J).

(iv) False
A stretched rubber band at rest has potential energy, not kinetic energy.

(v) True
Energy can change from one form to another, such as potential energy converting into kinetic energy.

2. Fill in the blanks.
(i) Work done = ______ × ______ (in the direction of force).
(ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.
(iii) The expression for kinetic energy of a body of mass m and velocity v is ______.
(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ______.
(v) Power is defined as the ______ at which work is done.

Answer
(i) Work done $=$ Force × Displacement (in the direction of force)
(ii) 1 joule of work is done when a force of 1 newton displaces an object by 1 metre in the direction of the force.
(iii) The expression for kinetic energy of a body of mass m and velocity v is $\mathbf{1} \boldsymbol{/} \mathbf{2} \mathbf{m v}^{\mathbf{2}}$.
(iv) The potential energy of an object of mass $m$ at a small height $h$ from the Earth's surface is $\mathbf{m g h}$.
(v) Power is defined as the rate at which work is done.

3. When a ball thrown upwards reaches its highest point, tick which of
the following statement(s) are correct?
(i) The force acting on the ball is zero.
(ii) The acceleration of the ball is zero.
(iii) Its kinetic energy is zero.
(iv) Its potential energy is maximum.

Answer:

The correct statements are:

(iii) Its kinetic energy is zero
(iv) Its potential energy is maximum

4. For each of the following situations, identify the energy transformation
that takes place:
(i) a truck moving uphill,
(ii) unwinding of a watchspring,
(iii) photosynthesis in green leaves,
(iv) water flowing from a dam,
(v) burning of a matchstick,
(vi) explosion of a fire cracker,
(vii) speaking into a microphone,
viii) a glowing electric bulb, and
(ix) a solar panel.

Answer

(i) Truck moving uphill: Kinetic energy changes into potential energy

(ii) Unwinding of a watch spring: Potential energy changes into kinetic energy

(iii) Photosynthesis in green leaves: Light energy changes into chemical energy

(iv) Water flowing from a dam: Potential energy changes into kinetic energy

(v) Burning of a matchstick: Chemical energy changes into heat and light energy

(vi) Explosion of a firecracker: Chemical energy changes into heat, light, and sound energy

(vii) Speaking into a microphone: Sound energy changes into electrical energy

(viii) A glowing electric bulb: Electrical energy changes into light and heat energy

(ix) A solar panel: Light energy changes into electrical energy

5. A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is $h=72.5 \mathrm{~m}$, acceleration due to gravity is $g=10 \mathrm{~m} \mathrm{~s}^{-2}$, and student's mass is $m=50 \mathrm{~kg}$.
(i) Find the gain in the potential energy if the student is lifted straight up to the top.
(ii) Find the gain in the potential energy when the student climbs the stairs to the same top.
(iii) What do you conclude about the dependence of the potential energy on the path taken?

Answer
Given:

$m=50 \mathrm{~kg}, \quad g=10 \mathrm{~m} / \mathrm{s}^2, \quad h=72.5 \mathrm{~m}$

(i) Gain in potential energy (elevator):

$
\text { P.E. }=m g h=50 \times 10 \times 72.5=36250 \mathrm{~J}
$

(ii) Gain in potential energy (stairs):

$
\text { P.E. }=m g h=50 \times 10 \times 72.5=36250 \mathrm{~J}
$

(iii) Conclusion: The gain in potential energy is the same in both cases. This shows that potential energy depends only on the initial and final positions (height), not on the path taken.

6. A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.

Answer:

Let the height of each floor be $h$.
The height of the 10th floor is $10 h$ and the height of the 20th floor is $20 h$.

Energy required
Work done $=m g h$
For lifting the mass to the 10th floor:

$W_1=m g(10 h)$

For lifting the mass to the 20th floor:

$
W_2=m g(20 h)
$

Thus,

$W_2=2 W_1$

So, the energy required to reach the 20 th floor is twice the energy required for the 10 th floor.

Power required
Power $=\frac{\text { Work }}{\text { Time }}$
Time taken to reach the 10th floor $=t$
Time taken to reach the 20th floor $=2 t$

$
P_1=\frac{W_1}{t}, \quad P_2=\frac{W_2}{2 t}
$

Since $W_2=2 W_1$,

$
P_2=\frac{2 W_1}{2 t}=\frac{W_1}{t}=P_1
$

7. Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at
which the flag is raised is doubled, how does the power requirement change? Explain your answers.

Answer: The energy required to raise a flag depends on the mass of the flag, the height of the flagpole, and the acceleration due to gravity (g). It is given by the expression mgh.
Raising the flag slowly or quickly does not change the amount of work done, because work depends only on the height and mass, not on time.

However, power depends on time. If the flag is raised faster, less time is taken, so more power is required. If the speed is doubled, the time taken becomes half, and therefore the power required becomes double.

8. A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction

Answer:

Mass of man $=60 \mathrm{~kg}$
Mass of scooter $=100 \mathrm{~kg}$
First day total mass: $m_1=60+100=160 \mathrm{~kg}$

Second day total mass: $m_2=60+40+100=200 \mathrm{~kg}$

The kinetic energy required to reach speed $v$ is given by:

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

Since speed $v$ is same in both cases, energy is directly proportional to mass.

$\frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{160}{200}=\frac{4}{5}$

The ratio of fuel used on the two days is 4 : 5.
More fuel is required on the second day because the total mass is greater.

9. On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.

Answer

For a seesaw to be balanced, the moments on both sides must be equal.

Moment=Force×Distance

Since the adult weighs twice the child, the adult must sit at half the distance from the fulcrum compared to the child.

$\mathrm{d}_1=2 \mathrm{~d}_2$

10. A ball of mass 2 kg is thrown up with a velocity of $20 \mathrm{~m} \mathrm{~s}^{-1}$.
(i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.
(ii) If the ball reaches a height of 19.4 m , how much work was done by air resistance (assume $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ ).

Answer
Given:
Mass $m=2 \mathrm{~kg}$, velocity $v=20 \mathrm{~m} / \mathrm{s}$, height $h=19.4 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2$
(i) Sign of work done by gravity
During upward motion: Gravity acts downward while displacement is upward, so work done by gravity is negative.
During downward motion:Gravity and displacement are in the same direction, so work done by gravity is positive.
(ii) Work done by air resistance

Initial kinetic energy:

$
\text { K.E. }=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times(20)^2=400 \mathrm{~J}
$
Potential energy at height $h$ :

$
\text { P.E. }=m g h=2 \times 10 \times 19.4=388 \mathrm{~J}
$

Loss of energy due to air resistance:

$\text { Work done by air resistance }=400-388=12 \mathrm{~J}
$

Since air resistance opposes motion, work done is negative.

11. A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m . If the block had a kinetic energy of 180 J when it was at 0 m , find the block's speed (i) at 0 m , and (ii) at 4 m . Does the block have negative acceleration in any portion of its motion?

Answer
Given:
Mass $m=10 \mathrm{~kg}$, Initial K.E. $=180 \mathrm{~J}$
(i) Speed at 0 m

$\begin{gathered}
\text { K.E. }=\frac{1}{2} m v^2 \\
180=\frac{1}{2} \times 10 \times v^2 \\
180=5 v^2 \\
v^2=\frac{180}{5}=36 \\
v=6 \mathrm{~m} / \mathrm{s}
\end{gathered}$

Work done (0 to 4 m)
From 0 to 1 m (triangle):

$
W_1=\frac{1}{2} \times 1 \times 50=25 \mathrm{~J}
$

From 1 to 3 m (rectangle):

$
W_2=2 \times 50=100 \mathrm{~J}
$
From 3 to 4 m (triangle):

$
W_3=\frac{1}{2} \times 1 \times 50=25 \mathrm{~J}
$

Total work $=25+100+25=150 \mathrm{~J}$

(ii) Speed at 4 m

$
\begin{gathered}
\text { Final K.E. }=180+150=330 \mathrm{~J} \\
\begin{array}{c}
330=\frac{1}{2} \times 10 \times v^2 \\
330=5 v^2 \\
v^2=\frac{330}{5}=66 \\
v=\sqrt{66} \approx 8.1 \mathrm{~m} / \mathrm{s}
\end{array}
\end{gathered}
$

(iii) Negative acceleration

From 3 m to 4 m , the force decreases but remains positive. So, acceleration decreases but is not negative.

12. The gravitational attraction on the surface of the Moon (lunar surface) is about $\frac{1}{6}$ th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?

Answer
On Earth, maximum height:

$h_E=\frac{u^2}{2 g_E}$

On Moon, maximum height: $h_M=\frac{u^2}{2 g_M}$

Given:

$g_M=\frac{g_E}{6}$

Calculation

$
\begin{gathered}
h_M=\frac{u^2}{2\left(g_E / 6\right)}=\frac{u^2}{2 g_E} \times 6 \\
h_M=6 h_E \\
h_M=6 \times 8=48 \mathrm{~m}
\end{gathered}
$

13. A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B . (ii) Calculate the kinetic energy of the car at A . (iii) State the work done by the brakes in bringing the car to a halt between B and C . (iv) What does the kinetic energy of the car transform into?

Answer

(i) Between A and B, the graph shows a horizontal line, so the car moves with constant speed ( $35 \mathrm{~m} / \mathrm{s}$ ). There is no acceleration.
(ii) Kinetic energy at $\mathbf{A}$

$
\begin{gathered}
\text { K.E. }=\frac{1}{2} m v^2 \\
=\frac{1}{2} \times 1000 \times(35)^2 \\
=500 \times 1225=612500 \mathrm{~J}
\end{gathered}
$

(iii) Work done by brakes (B to C)

Final velocity at $\mathrm{C}=0$

$
\text { Work done }=0-612500=-612500 \mathrm{~J}
$
Work done by brakes $=-\mathbf{6 1 2 5 0 0 ~ J}$
(iv) The kinetic energy is converted into heat energy due to friction in the brakes and tyres.

14. The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is $0 \mathrm{~m} \mathrm{~s}^{-1}$ and potential energy is 30 J . Calculate the velocity of the ball at $\mathrm{P}, \mathrm{Q}$ and R .

Answer
Given:
Mass $m=0.5 \mathrm{~kg}$
At point O , velocity $=0$ and potential energy $=30 \mathrm{~J}$
Total mechanical energy remains constant:

$
E=30 \mathrm{~J}
$

At P
From the graph, potential energy at $\mathrm{P}=20 \mathrm{~J}$

$
\begin{gathered}
\text { K.E. }=30-20=10 \mathrm{~J} \\
\qquad \begin{array}{c}
10=\frac{1}{2} \times 0.5 \times v^2 \\
10=0.25 v^2 \\
v^2=40 \\
v=\sqrt{40} \approx 6.32 \mathrm{~m} / \mathrm{s}
\end{array}
\end{gathered}
$

At Q
From the graph, potential energy at Q $=30 \mathrm{~J}$

$
\begin{aligned}
& \text { K.E. }=30-30=0 \\
& \qquad v=0 \mathrm{~m} / \mathrm{s}
\end{aligned}
$

At R
From the graph, potential energy at $\mathrm{R}=40 \mathrm{~J}$

$
\text { K.E. }=30-40=-10 \mathrm{~J}
$

This is not possible, so the ball cannot reach point $R$.

15. A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.
(i) Calculate the velocity of the coconut just before it hits the sand.
(ii) Assume that the average resistive force of sand is 3000 N and all of the coconut's energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume $g=10 \mathrm{~m} \mathrm{~s}^{-2}$.

Answer:

Given:
Mass $m=1.5 \mathrm{~kg}$, height $h=10 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2$, resistive force $F=3000 \mathrm{~N}$
(i) Velocity just before hitting the sand

$
\begin{gathered}
v^2=u^2+2 g h \\
v^2=0+2 \times 10 \times 10=200 \\
v=\sqrt{200}=10 \sqrt{2} \approx 14.14 \mathrm{~m} / \mathrm{s}
\end{gathered}
$

(ii) Depth of depression

Loss of potential energy = Work done against sand

$
\begin{gathered}
\text { P.E. }=m g h=1.5 \times 10 \times 10=150 \mathrm{~J} \\
\begin{array}{c}
\text { Work done }=F \times d=3000 \times d \\
3000 d=150 \\
d=\frac{150}{3000}=0.05 \mathrm{~m}
\end{array}
\end{gathered}
$

NCERT Class 9 Science Chapter 7 - Work, Energy, and Simple Machines: Think It Over Question Answer

This section provides clear and simple answers to the Think It Over questions of the chapter. It helps students develop deeper understanding and apply concepts in real-life situations.

1. What will be the magnitude of the velocity of the child at the bottom of the blue slide?

Answer: The magnitude of the velocity of the child at the bottom of the blue slide depends only on the height of the slide. As the child comes down, potential energy converts into kinetic energy. Hence, the child will have a certain velocity depending on the height from which they start.

2. Will two children of different masses reach the bottom of the same slide with the same velocity?

Answer: Yes, two children of different masses will reach the bottom with the same velocity, provided they start from the same height and friction is negligible. This is because velocity depends on height, not on mass.

3. Which of the slides will result in the largest magnitude of velocity for the child at its bottom?

Answer: The slide with the greatest vertical height will give the largest velocity at the bottom. If all slides have the same starting height, then the velocity at the bottom will be the same for all slides.

Class 9 NCERT Science Chapter 7 - Work, Energy, and Simple Machines Pause and Ponder Questions and Answers:

This section provides simple and clear answers to the Pause and Ponder questions of the chapter. It helps students think more deeply about the concepts and strengthen their understanding.

NCERT Science Class 9 Chapter 7 Pause and Ponder(Page 119)

1. In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?

Answer

No, the weightlifter is not doing any work on the barbell while holding it steady.

This is because work is done only when a force causes displacement. In this case, although the weightlifter is applying force to hold the barbell, there is no displacement, so no work is done according to physics.

2. Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c) - positive, negative or zero?

Answer:

The work done by friction is negative.

This is because friction always acts in the direction opposite to the motion of the coins. Since force and displacement are in opposite directions, the work done by friction is negative.

NCERT Class 9 Science Chapter 7 Pause and Ponder(Page 121)

3. When you pedal a bicycle on a flat road, your muscles supply energy. In
what forms does this muscular energy appear as you ride?

Answer: When a person pedals a bicycle, the muscular energy is converted mainly into the kinetic energy of the moving bicycle.

At the same time, some part of this energy is also converted into heat energy due to friction in the moving parts and a small amount into sound energy.

NCERT Science Class 9 Chapter 7 Pause and Ponder(Page 123)

4. Two objects A and B of mass m and 4 m have the same kinetic energy. What is the ratio of the
magnitude of velocities of A and B?

Answer:
Kinetic energy K.E. $=\frac{1}{2} m v^2$
For object A (mass $m$ ):

$
\text { K.E. }=\frac{1}{2} m v_A^2
$

For object B (mass $4 m$ ):

$
\text { K.E. }=\frac{1}{2}(4 m) v_B^2
$

Since kinetic energies are equal:

$
\begin{aligned}
\frac{1}{2} m v_A^2 & =\frac{1}{2}(4 m) v_B^2 \\
m v_A^2 & =4 m v_B^2 \\
v_A^2 & =4 v_B^2 \\
v_A & =2 v_B
\end{aligned}
$

5. Does the kinetic energy of an object which moves with constant velocity change with its position?

Answer:
No, the kinetic energy of an object moving with constant velocity does not change with its position.
Kinetic energy depends only on the mass and velocity of the object. Since the velocity remains constant, the kinetic energy also remains constant.

Class 9 NCERT Science Chapter 7 Pause and Ponder(Page 126)

6. Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?

Answer

No, the potential energy does not change when the object moves with constant velocity in the horizontal direction. This is because its height remains the same.

However, when the object is raised in the vertical direction, its potential energy increases. This is because potential energy depends on height above the ground.

NCERT Science Chapter 7 Class 9 Pause and Ponder(Page 129)

7. For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.

Answer
At the top (point A), the ball has only potential energy:

$
\text { P.E. }=m g h
$

As the ball falls, potential energy converts into kinetic energy.
Just before hitting the ground, height becomes zero, so potential energy becomes zero and all the energy is kinetic energy.

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

From the equation of motion:

$
\begin{gathered}
v^2=2 g h \\
\text { K.E. }=\frac{1}{2} m(2 g h)=m g h
\end{gathered}
$

8. You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points $\mathrm{A}, \mathrm{B}$ and C . Why do subsequent points, such as $\mathrm{C}, \mathrm{D}$ and E , usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?

Answer

At point A, the ball is at a higher position, so it has maximum potential energy and minimum kinetic energy.

At point B, the ball is lower, so some potential energy converts into kinetic energy. Thus, potential energy decreases and kinetic energy increases. At point C, the ball again rises to a height, so kinetic energy decreases and potential energy increases.

The subsequent points like C, D, and E have lower heights compared to the previous ones because some energy is lost due to friction and air resistance. This energy is converted into heat and sound, so the ball cannot reach the same height again.

Class 9 NCERT Science Chapter 7 Pause and Ponder(Page 132)

9. Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?

Answer

Roads on hills are built in a winding manner with gentle slopes to make it easier for vehicles to climb.

If the road were straight and steep, a larger force would be required to move upward. By increasing the path length with a gentle slope, the same height is reached with less force, making the journey safer and more comfortable.

10. To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.

Answer: Climbing an inclined ladder is easier because the force required is less.

When we climb an inclined ladder, we cover a longer distance to reach the same height, so less force is needed. In a vertical ladder, the height is gained in a shorter distance, so more force is required.

NCERT Science Chapter 7 Class 9 Pause and Ponder(Page 135)

11. Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?

Answer:

It is easier to open the lid of a can using a spoon because the spoon acts as a lever.

The edge of the can acts as the fulcrum, and when force is applied at the handle of the spoon, it produces a larger force at the lid. This makes it easier to lift the lid with less effort.

12. Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?

Answer:

Scissors act as a lever. When the object is placed closer to the fulcrum, the effort arm becomes longer compared to the load arm. This increases the mechanical advantage, so a greater force is applied at the cutting edges. Hence, it becomes easier to cut hard objects with less effort.

13. Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.

Answer:
All real machines slow down and eventually stop because some part of their energy is always lost due to friction and air resistance.
This lost energy is converted into forms like heat and sound, which cannot be fully used again to do work. As a result, the total useful mechanical energy keeps decreasing, and the machine cannot keep moving forever.
Hence, a perpetual machine is not possible because energy is always dissipated in real systems.

NCERT Class 9 Science Chapter 7 - Work, Energy, and Simple Machines Activity Question Answer:

NCERT Class 9 Science Chapter 7 Activity 7.1: Let us investigate

1. Take a heavy ball and a large container filled with loose sand.
2. Raise the ball over the sand bed to a height of about 1 m and drop it (Fig. 7.17). Is a depression created in the sand? Why does the ball create a depression?

3. Now, raise the ball to the height of 2 m and release it at a slightly different position over the sand bed such that the depressions do not overlap. Repeat this step one more time. Compare the depths of the depressions. Is there any difference? In which case is the depression deepest and in which case the shallowest?

Answer

When the ball is dropped on the sand, a depression is formed. This is because the ball has potential energy due to its height, which converts into kinetic energy as it falls. On hitting the sand, this energy is used to do work, pushing the sand particles aside and creating a depression.

When the ball is dropped from a greater height, it has more energy at the time of impact. Therefore, the depression formed is deeper. The depression is shallowest when the ball is dropped from the lowest height and deepest when dropped from the greatest height.

NCERT Class 9 Science Chapter 7 Activity 7.2: Let us experiment

1. Set up a simple pendulum as you have learnt in Grade 7.
2. Paste a white sheet of paper on a wall behind the pendulum. Draw a horizontal line above the position
of the bob when it is not oscillating (Fig. 7.20).

3. Take the bob to one side to a point P, which is at the level of the horizontal line and let it go. Observe it at the extreme points of the first couple of oscillations. Does the bob almost reach the level of the horizontal line?

Answer
When the bob is released from point P, it starts oscillating. At the extreme points, the bob almost reaches the same horizontal level again.
This happens because the potential energy at the extreme position converts into kinetic energy at the lowest point and then back into potential energy. Due to small energy losses (like air resistance), it may not reach exactly the same height, but it comes very close.

Class 9 NCERT Science Chapter 7 Activity 7.3: Let us experiment

1. Take a smooth plank (say a cardboard piece) about 1.5 m long, a toy car (or the cart that you used in Activity 6.3) and a spring balance. Attach the spring balance to the cart. Arrange an elevated surface, such as the top of a low stool or a pile of books, at about 0.5 m height from the floor.
2. First, lift the cart vertically, slowly and steadily, from the floor to the top of the pile of books or stool, and note the reading of the spring balance scale. This reading indicates the force required to lift the cart vertically which is equal to the weight of the cart.
3. Next, place the plank against the top of the pile of books or stool as shown in Fig. 7.27a. Pull the cart along the plank slowly and steadily. Is the reading of the spring balance (the force required) smaller than that of step 2 ?

4. Now, reduce the angle between the plank and the base as shown in Fig. 7.27b, and repeat the step 3. Observe how the force required changes as the plank becomes less steep.

Answer

When the cart is lifted vertically, the force required is equal to its weight.

When the cart is pulled along the inclined plank, the force required is less than its weight. As the plank becomes less steep, the force required decreases further.

This happens because an inclined plane increases the distance over which the work is done, thereby reducing the force needed to lift the object to the same height.

NCERT Solution for Class 9 Chapter 7 Activity 7.4: Let us investigate

1. Take a 30 cm long scale, a pencil, $2-3$ erasers and a stapler (or a similar object).
2. Place the scale over the pencil such that the pencil is closer to one end of the scale as shown in Fig. 7.31. On the end of the scale closer to the pencil, place the stapler.
3. On the other end of the scale, place one eraser. Does the stapler lift up? If not, add one more eraser.

Answer

In this activity, the stapler is lifted even though erasers are lighter. This happens because the scale acts as a lever with the pencil as the fulcrum.

When the erasers are placed farther from the fulcrum, they produce a greater turning effect (moment) compared to the stapler placed near the fulcrum. Hence, a smaller force can lift a heavier object.

Class 9 Chapter 7 Activity 7.5: Let us experiment

1. Take a long scale ( 50 cm or larger), a piece of string, two paper cups (to act as pans), adhesive tape or a piece of thread, and identical coins (to act as weights).
2. Tie the string tightly around the scale at its midpoint. This string will act as the fulcrum. Hang the scale from this string using a stand or hook, so that it can swing freely. This scale will now act as a beam (Fig. 7.33).
3. Fix paper cups to both ends of the beam using thread. These cups act as the pans of a balance. Check whether the beam is levelled. If it is tilted, adjust the hanging points of the pans until both sides balance equally.
4. Place 1 coin in the left pan (call it effort) and 1 identical coin in the right pan (call it load). Observe that the beam stays horizontal.
5. Add one more coin to the right pan, so that it contains 2 coins. The beam tilts. Move the heavier pan closer to the centre of the beam to balance the beam. Measure its distance from the centre.
6. Repeat step 5 with 4 coins and then 8 coins in the right pan. Each time, note its distance from the centre that balances the beam.
7. Record all observations and measurements, and complete the Table 7.1 by adding more rows.

Answer
In this activity, the beam balances when the product of number of coins and their distance from the fulcrum is equal on both sides.

This shows that for a lever:

$
\text { Effort × Effort arm }=\text { Load × Load arm }
$

Table 7.1: Number of coins and distance from fulcrum

Conclusion
The product of number of coins and distance from the fulcrum is equal on both sides:

$
n_1 \times L_1=n_2 \times L_2
$

This verifies the principle of moments.

NCERT Solution For Class 9 Science Chapter 7: Topics

This section gives you a simple list of all the important topics in the chapter. It helps you understand what to study and makes revision quick and easy.

7.1 Work Done by a Constant Force

• 7.1.1 When is work done equal to zero?
• 7.1.2 Positive and negative work done

7.2 The Work-Energy Theorem
7.3 Forms of Energy
7.4 Mechanical Energy

• 7.4.1 Kinetic energy
• 7.4.2 Potential energy
• 7.4.3 Conservation of mechanical energy

7.5 Power
7.6 Simple Machines

• 7.6.1 Pulley
• 7.6.2 Inclined plane
• 7.6.3 Lever