NCERT Solutions for Class 9 Science Exploration Chapter 6

Force is a push or pull that can change the state of rest, speed, direction, or shape of an object.
Newton’s laws of motion explain how balanced forces, unbalanced forces, friction, mass, acceleration, and action-reaction pairs affect motion.

NCERT Solutions for Class 9 Science Exploration Chapter 6 help students understand How Forces Affect Motion through force diagrams, daily-life examples, Newton’s laws, and numerical problems. This chapter continues from motion by asking what causes changes in velocity and why some objects stay at rest, move with constant velocity, accelerate, or stop. Students should focus on the difference between balanced and unbalanced forces, the role of friction, and the formulas $F = ma$ and $F = mg$. These NCERT Class 9 Science Solutions cover Pause and Ponder questions and all Revise, Reflect, Refine exercise answers from Class 9 Science Exploration Chapter 6.

Key Takeaways

• Force: Force has both magnitude and direction, and its SI unit is newton, written as $N$.
• Balanced forces: Balanced forces do not change the state of rest or uniform motion of an object.
• Unbalanced forces: A non-zero net force changes velocity and produces acceleration.
• Newton’s laws: The three laws connect inertia, force, mass, acceleration, and action-reaction pairs.

NCERT Solutions for Class 9 Science Exploration Chapter 6 Structure 2026

NCERT Solutions for Class 9 Science Exploration Chapter 6 In-Text Questions

Class 9 Science Exploration Chapter 6 uses examples like lifting a barbell, arm-wrestling, toy cars, airbags, walking, rowing a canoe, and rocket motion to explain force and motion. The answers below follow the textbook concepts directly.

Q1. Which forces act on a barbell lifted by a weightlifter?

Question: A weightlifter lifts a barbell. List two forces acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?

Answer: The two forces are the gravitational force acting downward and the muscular force applied by the weightlifter acting upward.

Explanation:
If the barbell is steady, its velocity is not changing. This means the net force on it is zero.

So, the upward force applied by the weightlifter balances the downward gravitational force.

Therefore, the forces are balanced when the barbell is held steady.

Q2. Are forces balanced in an arm-wrestling match when the arms tilt?

Answer: No, the forces are not balanced when the arms tilt.

Explanation:
If the arms move or tilt in one direction, there must be a non-zero net force or turning effect in that direction.

The player whose applied force is more effective in the direction of tilt has exerted the larger force at that instant.

Balanced forces would not produce such motion.

Q3. An object moves with constant velocity. Is there a net force acting on it?

Answer: No, the net force acting on an object moving with constant velocity is zero.

Explanation:
According to Newton’s first law, an object continues to move with constant velocity unless a net force acts on it.

Constant velocity means no change in speed or direction. So, acceleration is zero.

Using Newton’s second law:

$F = ma$

If $a = 0$, then:

$F = 0$

Q4. What happens when no net force acts on an object?

Question: Suppose no net force is acting on an object. Which situations are possible?

(i) Object remains at rest if at rest.
(ii) Object keeps moving with a constant velocity if already moving.
(iii) Object is moving with a constant acceleration.

Answer: The possible situations are (i) and (ii).

Explanation:
If no net force acts, acceleration is zero.

An object at rest remains at rest. An object already moving continues to move with constant velocity.

The third statement is incorrect because constant acceleration requires a non-zero net force.

Q5. Give an example where net force is zero even though forces act on an object.

Answer: A book lying on a table has forces acting on it, but the net force is zero.

Explanation:
The gravitational force pulls the book downward. The table applies a normal force upward.

These two forces are equal in magnitude and opposite in direction. They balance each other.

So, the book remains at rest because the net force is zero.

Q6. What is the net force on a toy car moving with constant velocity?

Question: A toy car of mass $100 , g$ is moving with a constant velocity of $0.5 , m , s^{-1}$. What is the net force acting on it?

Answer: The net force is $0 , N$.

Explanation:
The car is moving with constant velocity. Therefore, its acceleration is zero.

Using:

$F = ma$

$F = 0.1 \times 0$

$F = 0 , N$

The toy car may be moving, but no net force is acting on it.

Q7. Which child on a swing needs a larger force for the same acceleration?

Answer: The child with greater mass needs a larger force.

Explanation:
Newton’s second law states:

$F = ma$

For the same acceleration, force is directly proportional to mass.

So, a heavier child requires a larger force to get the same initial acceleration as a lighter child.

Q8. How does bubble wrap or hay protect glass items?

Answer: Bubble wrap or hay increases the time during which the glass item stops during impact.

Explanation:
When a glass item falls or receives a sudden jerk, its velocity changes quickly. If the stopping time is very small, the force is large.

Soft packing material increases the stopping time and reduces acceleration.

Since:

$F = ma$

smaller acceleration means smaller force on the glass item. This reduces the chance of damage.

Q9. Why does a fireperson struggle while holding a pipe issuing water?

Answer: A fireperson struggles because water rushing out of the pipe exerts an equal and opposite force on the pipe.

Explanation:
When water is pushed forward through the pipe, the water also pushes the pipe backward.

This follows Newton’s third law of motion.

The backward reaction force can be large because water comes out at high speed. The fireperson must apply force to hold the pipe steady.

Q10. How can a spacecraft change velocity in space where gravitational force is negligible?

Answer: A spacecraft can change its velocity by firing its engines and ejecting gases in the opposite direction.

Explanation:
In space, the spacecraft does not need air or ground to push against. Its engine expels gases in one direction.

According to Newton’s third law, the gases push the spacecraft in the opposite direction.

If the engine fires opposite to the direction of motion, the spacecraft speeds up. If it fires in the direction of motion, the spacecraft slows down.

NCERT Solutions for Class 9 Science Exploration Chapter 6 Exercise Questions

The Revise, Reflect, Refine section includes conceptual MCQs, force diagrams, graph-based questions, Newton’s law applications, and numerical problems. These Class 9 Science Chapter 6 exercise solutions are written in direct answer format.

Q1. A table is moved with constant velocity using a horizontal force F. What is the frictional force?

Answer: The frictional force is equal in magnitude to $F$ and acts in the opposite direction.

Explanation:
The table moves with constant velocity, so acceleration is zero.

Using Newton’s second law:

$F_{net} = ma = 0$

Therefore, the applied force is balanced by friction.

If applied force is $F$, then frictional force is also $F$ in the opposite direction.

Q2. Choose the correct options for a ball moving on a smooth frictionless surface.

Answer:
(i) The velocity of the ball will remain the same.
(ii) The magnitude of velocity will increase.
(iii) The magnitude of velocity will decrease.

Explanation:
On a frictionless surface, if no net force acts, the ball continues with constant velocity.

If net force acts in the direction of motion, acceleration is in the same direction, so speed increases.

If net force acts opposite to motion, acceleration is opposite to velocity, so speed decreases.

Q3. Which statement is correct for blocks P and Q?

Question: Two forces of $4 , N$ and $5 , N$ act in opposite directions on block P. Block Q is moving with constant velocity. Choose the correct statement.

Answer: The correct option is (i) P experiences a net force and Q does not experience a net force.

Explanation:
For block P:

$Net , force = 5 , N - 4 , N = 1 , N$

So, block P experiences a net force.

Block Q is moving with constant velocity, so acceleration is zero. Hence, net force on Q is zero.

Q4. Find the net force on a snake boat.

Question: In a snake boat race, 95 oarsmen row backwards to propel the boat forward, while 5 row in the opposite direction. Each applies $200 , N$. Find the net force.

Answer: The net force on the boat is $18000 , N$ forward.

Explanation:
Force by 95 oarsmen:

$95 \times 200 = 19000 , N$

Force by 5 oarsmen in the opposite direction:

$5 \times 200 = 1000 , N$

Net force:

$19000 - 1000 = 18000 , N$

So, the net force is $18000 , N$ in the forward direction.

Q5. What happens when a net force acts on an object?

Answer: The correct option is (iv) in the direction of force, with acceleration proportional to the force acting on the object.

Explanation:
According to Newton’s second law, acceleration occurs in the direction of the net force.

Also:

$a = \frac{F}{m}$

For a given mass, acceleration is directly proportional to net force.

Q6. Which object has a net force acting on it from the position-time graphs?

Answer: A net force acts on the object whose position-time graph is curved.

Explanation:
A straight horizontal position-time graph shows rest. A straight sloping position-time graph shows constant velocity.

In both cases, acceleration is zero and net force is zero.

A curved position-time graph shows changing velocity, which means acceleration is present. Therefore, net force acts on that object.

Q7. What happens to the boat when a sailor jumps forward?

Answer: The boat moves backward.

Explanation:
When the sailor jumps forward, they push the boat backward with their feet.

According to Newton’s third law, the boat pushes the sailor forward with an equal and opposite force.

The backward force on the boat makes it move in the direction opposite to the sailor’s jump.

Q8. Why is a landing mat or sand bed used in high jump?

Answer: A landing mat or sand bed reduces the force acting on the athlete during landing.

Explanation:
When the athlete lands, their velocity becomes zero.

A soft mat increases the time taken to stop the athlete’s body. This reduces the magnitude of acceleration.

Using:

$F = ma$

smaller acceleration means smaller force on the athlete. This reduces injury.

Q9. What force do loaded and empty hand carts exert during collision?

Answer: The correct option is (iv) the loaded cart and the empty cart both exert an equal magnitude of force on each other.

Explanation:
According to Newton’s third law, when two objects interact, they exert equal and opposite forces on each other.

The loaded cart has greater mass, so its acceleration may be smaller. The empty cart may accelerate more.

However, the forces during interaction are equal in magnitude and opposite in direction.

Q10. Plot the force-mass graph from the acceleration-mass graph.

Answer: The force-mass graph will be a horizontal straight line because the force is constant.

Explanation:
From the graph, acceleration decreases as mass increases.

Using:

$F = ma$

If values such as $m = 1 , kg$ and $a = 10 , m , s^{-2}$ are taken:

$F = 1 \times 10 = 10 , N$

For $m = 2 , kg$ and $a = 5 , m , s^{-2}$:

$F = 2 \times 5 = 10 , N$

So, the force remains $10 , N$ for all masses.

The force-mass graph is a horizontal line at $F = 10 , N$.

Q11. Calculate force from velocity-time graph for a 10 kg object.

Answer: The force acting on the object is $25 , N$.

Explanation:
From the graph, velocity increases from $0$ to $20 , m , s^{-1}$ in $8 , s$.

Acceleration:

$a = \frac{v-u}{t}$

$a = \frac{20 - 0}{8}$

$a = 2.5 , m , s^{-2}$

Mass:

$m = 10 , kg$

Force:

$F = ma$

$F = 10 \times 2.5$

$F = 25 , N$

The force acts in the direction of motion.

Q12. Estimate the stopping force on a bullet.

Question: A bullet of mass $50 , g$ moving at $100 , m , s^{-1}$ stops after penetrating $50 , cm$ into a wooden block. Estimate the stopping force.

Answer: The stopping force is $500 , N$ opposite to the motion.

Explanation:
Convert units:

$m = 50 , g = 0.05 , kg$

$u = 100 , m , s^{-1}$

$v = 0 , m , s^{-1}$

$s = 50 , cm = 0.5 , m$

Use:

$v^2 = u^2 + 2as$

$0^2 = 100^2 + 2 \times a \times 0.5$

$0 = 10000 + a$

$a = -10000 , m , s^{-2}$

Force:

$F = ma$

$F = 0.05 \times (-10000)$

$F = -500 , N$

The negative sign means the force acts opposite to the bullet’s motion.

Q13. Calculate time of contact between footballer’s foot and ball.

Question: A football of mass $0.4 , kg$ is kicked to a speed of $108 , km , h^{-1}$. Force applied is $800 , N$. Find contact time.

Answer: The time of contact is $0.015 , s$.

Explanation:
Convert speed:

$108 , km , h^{-1} = 30 , m , s^{-1}$

Given:

$m = 0.4 , kg$

$u = 0 , m , s^{-1}$

$v = 30 , m , s^{-1}$

$F = 800 , N$

Using:

$F = ma$

$a = \frac{F}{m}$

$a = \frac{800}{0.4} = 2000 , m , s^{-2}$

Now:

$v = u + at$

$30 = 0 + 2000t$

$t = \frac{30}{2000}$

$t = 0.015 , s$

Q14. Find distance travelled before stopping on a rough patch.

Question: An object of mass $2 , kg$ moves at $10 , m , s^{-1}$. Friction is $7 , N$, and an additional opposing force is $3 , N$. Find distance before rest.

Answer: The object travels $10 , m$ before coming to rest.

Explanation:
Total opposing force:

$7 + 3 = 10 , N$

Mass:

$m = 2 , kg$

Acceleration:

$a = \frac{F}{m}$

$a = \frac{10}{2} = 5 , m , s^{-2}$ opposite to motion

So:

$a = -5 , m , s^{-2}$

Given:

$u = 10 , m , s^{-1}$

$v = 0 , m , s^{-1}$

Use:

$v^2 = u^2 + 2as$

$0^2 = 10^2 + 2 \times (-5) \times s$

$0 = 100 - 10s$

$s = 10 , m$

Q15. Find acceleration when trolley and harrow are pulled together.

Question: A tractor pulls a harrow of mass $m_1$ with force $F$, producing acceleration $a_1$. It pulls a trolley of mass $m_2$ with the same force $F$, producing acceleration $a_2$. Find acceleration when trolley and harrow are pulled together with force $F$.

Answer: The resulting acceleration is:

$a = \frac{a_1a_2}{a_1 + a_2}$

Explanation:
Using Newton’s second law:

$F = m_1a_1$

So:

$m_1 = \frac{F}{a_1}$

Also:

$F = m_2a_2$

So:

$m_2 = \frac{F}{a_2}$

When both are pulled together:

$total , mass = m_1 + m_2$

$a = \frac{F}{m_1 + m_2}$

Substitute values:

$a = \frac{F}{\frac{F}{a_1} + \frac{F}{a_2}}$

$a = \frac{F}{F(\frac{1}{a_1} + \frac{1}{a_2})}$

$a = \frac{1}{\frac{1}{a_1} + \frac{1}{a_2}}$

$a = \frac{a_1a_2}{a_1 + a_2}$

Q16. Why does a compass needle move but the bar magnet does not?

Answer: The compass needle moves because it has much smaller mass and is free to rotate.

Explanation:
According to Newton’s third law, the bar magnet and compass needle exert equal and opposite magnetic forces on each other.

However, acceleration depends on mass:

$a = \frac{F}{m}$

The compass needle has small mass and is mounted to rotate easily. So, it shows noticeable motion.

The bar magnet has much larger mass and may be held by hand or placed firmly. Its acceleration is too small to notice.

NCERT Solutions for Class 9 Science Exploration

Topics Covered in NCERT Solutions for Class 9 Science Exploration Chapter 6

Class 9 Science Exploration Chapter 6 covers how forces affect rest, motion, acceleration, and interaction between objects. The chapter connects daily experiences with Newton’s laws of motion.

• Meaning of force
• Magnitude and direction of force
• SI unit of force
• Measuring force using spring balance
• Balanced and unbalanced forces Class 9
• Net force Class 9 Science
• Force of friction Class 9
• Normal force and gravitational force
• Newton’s first law Class 9
• Inertia and constant velocity
• Newton’s second law Class 9
• Relation between force, mass, and acceleration
• Formula $F = ma$
• Gravitational force formula $F = mg$
• Newton’s third law Class 9
• Action and reaction forces
• Walking, rowing, rocket launch, and recoil examples
• Forces acting on a system of objects

Important Concepts in NCERT Solutions for Class 9 Science Exploration Chapter 6

How Forces Affect Motion Class 9 requires students to identify the net force and connect it with acceleration. These concepts are important for both conceptual questions and numericals.

Important Formulas in NCERT Solutions for Class 9 Science Exploration Chapter 6

Class 9 Science Chapter 6 solutions use formulas from force, mass, acceleration, weight, and motion. Students should write units correctly while solving numericals.