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Chapter 6 How Forces Affect Motion Class 9th Science (Exploration) ncert solution

Class 9 · Science (Exploration) · Chapter 6: How forces affect motion · All Board · ENGLISH · 17 views

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Grade 9 · Science · Curiosity

How Forces Affect Motion

Chapter 6 — full, step-by-step solutions for every solved example, In-Text “Pause & Ponder”, and end-of-chapter exercise question, with diagrams and clean maths.

Newton's 3 Laws Friction Balanced & Unbalanced Forces F = ma
1

Solved Examples

The worked examples from the chapter, fully explained.

6.1
Two forces of 10 N and 6 N are acting on a block lying on a table (Fig. 6.6). Find the magnitude and direction of the net force in each case.
10 N6 N(a)10 N6 N(b)6 N10 N(c)
Fig. 6.6: Two forces acting on a block in three different ways
ICON Solution

Forces along the same straight line simply add when they point the same way and subtract when they point in opposite ways; the net force then points along the larger force.

  • (a) Both forces point right (same direction): $F_{net}=10\text{ N}+6\text{ N}=16\text{ N}$ towards the right.
  • (b) 10 N right, 6 N left: $F_{net}=10\text{ N}-6\text{ N}=4\text{ N}$ towards the right.
  • (c) 6 N right, 10 N left: $F_{net}=10\text{ N}-6\text{ N}=4\text{ N}$ towards the left.
Answer(a) 16 N right  •  (b) 4 N right  •  (c) 4 N left
6.2
A person exerts a force on a moving box in the forward direction that is equal to the force of friction between the box and the floor. Will the box keep moving or stop after some time?
ICON Solution

The applied force acts forward; friction acts backward with the same magnitude. These two forces are equal and opposite, so they balance each other and the net force is zero.

By Newton's first law, an object in motion with zero net force continues to move with constant velocity. So the box neither speeds up nor slows down.

AnswerThe box continues moving with constant velocity (it does not stop).
6.3
Draw the (i) position–time and (ii) velocity–time graphs for an object on which no net force is acting.
ICON Solution

If no net force acts, acceleration is zero, so there are only two possibilities: the object is at rest, or it moves with constant velocity.

TimePosition0At restTimeVelocity0At restTimePosition0Constant velocityTimeVelocity0Constant velocity
At rest (Fig. 6.15) and moving with constant velocity (Fig. 6.16)
  • At rest: position stays the same → position–time graph is a horizontal line; velocity is zero → velocity–time graph lies along the time axis.
  • Constant velocity: position increases uniformly → position–time graph is a straight inclined line; velocity is unchanging → velocity–time graph is a horizontal line above the axis.
6.4
A weightlifter holds a barbell with 10 kg fixed on each side of the bar (Fig. 6.8). The bar itself has a mass of 10 kg. How much force must she apply to keep the barbell steady? (Take $g = 9.8\ \text{m s}^{-2}$)
Weightlifter holding a barbell steady
Fig. 6.8: A weightlifter lifting a barbell
ICON Solution
  1. Find total mass$m = 10 + 10 + 10 = 30\ \text{kg}$
  2. Weight (downward gravitational force)$F = mg = 30 \times 9.8 = 294\ \text{N}$ (downwards)
  3. Apply Newton's first lawTo keep it steady (net force = 0), she must push up with a force equal to the weight.
Answer294 N, directed upward.
6.5
A student pushes a stationary block of 25 kg on a horizontal floor. The maximum force of friction opposing motion is 50 N. Find the displacement of the block in 2 s if it is pushed with a constant force of (i) 50 N and (ii) 55 N.
ICON Solution

(i) Push = 50 N. The applied force exactly equals friction (50 N), so the forces are balanced and the net force is zero. The block stays still.

(i)Displacement = 0 (block remains stationary).

(ii) Push = 55 N.

  1. Net force$F_{net} = 55 - 50 = 5\ \text{N}$ (forward)
  2. Acceleration (Newton's 2nd law)$a = \dfrac{F}{m} = \dfrac{5}{25} = 0.2\ \text{m s}^{-2}$
  3. Displacement in 2 sUsing $s = ut + \tfrac{1}{2}at^2$ with $u=0$:
    $s = 0 + \tfrac{1}{2}\times 0.2 \times (2)^2 = 0.4\ \text{m}$
(ii)0.4 m in the forward direction.
6.6
A sports car of mass 1500 kg moves towards the east; its velocity–time graph is shown (Fig. 6.21). Calculate the force on the car during (i) 0–5 s, (ii) 5–10 s, (iii) 10–15 s.
Time (s)Velocity (m s⁻¹)0510152.557.510
Fig. 6.21: Velocity–time graph for a sports car
ICON Solution

(i) 0–5 s — line slopes up, so constant acceleration with $u=0$, $v=10\ \text{m s}^{-1}$, $t=5\ \text{s}$:

  1. Acceleration$v = u + at \Rightarrow 10 = 0 + a(5) \Rightarrow a = 2\ \text{m s}^{-2}$
  2. Force$F = ma = 1500 \times 2 = 3000\ \text{N}$ towards the east

(ii) 5–10 s — line is parallel to the time axis ⇒ constant velocity ⇒ $a=0$ ⇒ F = 0 (no force).

(iii) 10–15 s — line slopes down, with $u=10$, $v=0$, $t=5\ \text{s}$:

  1. Acceleration$0 = 10 + a(5) \Rightarrow a = -2\ \text{m s}^{-2}$
  2. Force$F = 1500 \times (-2) = -3000\ \text{N}$
Answer(i) 3000 N east  •  (ii) 0 N  •  (iii) 3000 N west (negative sign = opposite to motion).
6.7
The Earth and a fruit pull on each other with equal and opposite gravitational forces (Fig. 6.33). Then why does the fruit fall towards the Earth while the Earth does not seem to move towards the fruit?
A fruit on a tree, the Earth and fruit attract each other
Fig. 6.33: Earth and fruit exert equal and opposite forces
ICON Solution

The forces on the two are equal in magnitude. But the resulting acceleration depends on mass, since $a = \dfrac{F}{m}$.

The Earth's mass is enormous compared with the fruit, so for the same force its acceleration is extremely tiny — far too small to notice. The fruit, being very light, gets a large acceleration and visibly falls.

AnswerEqual forces, but the Earth's huge mass gives it a negligible acceleration, while the light fruit accelerates noticeably.
6.8
A 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N, and the gun recoils. Find the magnitudes of the initial accelerations of the bullet and the gun.
ICON Solution

By Newton's third law, the recoil force on the gun is also 2 N.

  1. Acceleration of the gun$a_{gun} = \dfrac{F}{m_{gun}} = \dfrac{2}{5} = 0.4\ \text{m s}^{-2}$
  2. Acceleration of the bullet$a_{bullet} = \dfrac{F}{m_{bullet}} = \dfrac{2}{0.1} = 20\ \text{m s}^{-2}$

The forces are equal, but the bullet (smaller mass) accelerates far more than the gun.

AnswerGun: 0.4 m s⁻²  •  Bullet: 20 m s⁻²
2

In-Text Questions

“Pause and Ponder” questions appearing within the chapter.

1
A weightlifter lifts a barbell (Fig. 6.8). List two forces acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?
Weightlifter holding a barbell
Fig. 6.8: A weightlifter lifting a barbell
ICON Solution

Two forces act on the barbell:

  • The gravitational force (weight) of the barbell, acting downward.
  • The upward force applied by the weightlifter through her hands.

When the barbell is held steady, it is at rest (zero acceleration), so the net force on it must be zero. Hence the two forces are equal in magnitude and opposite in direction.

AnswerYes — they are balanced (equal up-force and down-weight, net force zero).
2
Two players R and S are arm-wrestling (Fig. 6.9). At the instant the arms tilt towards the front (out of the page, towards you), are the forces exerted by the players balanced? If not, who exerts the larger force?
Arm-wrestling match between R and S
Fig. 6.9: Arm-wrestling match (R on left, S on right)
ICON Solution

If the forces were balanced, the clasped hands would stay perfectly still. Because the arms are tilting (moving) towards the front, the hands are accelerating — so a net force acts and the forces are not balanced.

The net force, and therefore the motion, is in the direction of the larger push. So the player who is pushing the joined hands in the direction of the tilt (towards the front / out of the page) is exerting the larger force.

AnswerNot balanced. The player pushing the hands towards the front (the winning direction) exerts the larger force.
3
An object is moving with a constant velocity. Is there a net force acting upon it?
ICON Solution

Constant velocity means no change in speed or direction, so the acceleration is zero. By Newton's first (and second) law, zero acceleration ⇒ zero net force.

AnswerNo. The net force on the object is zero.
4
Suppose no net force acts on an object. Which of these are possible?
  • Object remains at rest if at rest.
  • Object keeps moving with a constant velocity if already moving.
  • Object is moving with a constant acceleration.
ICON Solution

No net force means zero acceleration. So an object stays at rest, or keeps its constant velocity. Any acceleration (including a constant one) requires a non-zero net force — so (iii) is impossible.

Answer(i) and (ii) are possible; (iii) is not possible.
5
In the real world it is hard to find an object with no forces acting on it, yet by applying additional forces the net force can be made zero. Explain with an example.
ICON Solution

Several real forces can cancel out so that their combined (net) effect is zero.

Example: A book lying on a table. Gravity pulls it down with its weight, while the table pushes up on it with an equal normal force. These balance, so the net force is zero and the book stays at rest.

Another example: Pushing a box across the floor at a steady speed — the forward push exactly balances the backward friction, giving zero net force and constant velocity.

AnswerBalancing extra forces (e.g. normal force vs weight, or push vs friction) can make the net force zero.
6
A toy car of mass 100 g moves with a constant velocity of $0.5\ \text{m s}^{-1}$. What is the net force on it?
ICON Solution

Constant velocity ⇒ zero acceleration. By $F = ma$, with $a = 0$:

$F = m \times 0 = 0\ \text{N}$

(The mass value is not even needed once we know the velocity is constant.)

AnswerNet force = 0 N.
7
Two children of different masses sit on identical swings. To give both the same initial acceleration, for which child must you apply a larger force? Explain why.
ICON Solution

From Newton's second law, $F = ma$. For the same acceleration $a$, the required force is directly proportional to the mass.

So the heavier child needs a larger force, because more force is needed to give a larger mass the same acceleration.

AnswerThe child with the larger mass (heavier child) needs the larger force.
8
How does packing glass items in bubble wrap or hay protect them from damage during transport?
ICON Solution

During a jolt or collision, the glass has to lose its momentum. Bubble wrap or hay increases the time over which the glass comes to rest.

A longer stopping time means a smaller deceleration. Since force $= \dfrac{\text{change in momentum}}{\text{time}}$ (i.e. $F = ma$), a smaller acceleration means a smaller force acts on the glass — reducing the chance of breaking.

This is the same idea behind a cricketer pulling their hands back while catching, airbags, and high-jump landing mats.
AnswerThe soft padding lengthens the impact time, lowering the force on the glass, so it is less likely to break.
9
Why does a fireperson sometimes struggle when holding a pipe that is issuing water?
ICON Solution

The pipe pushes a large stream of water forward (out of the nozzle). By Newton's third law, the water pushes back on the pipe with an equal and opposite force.

This strong backward reaction (recoil/thrust) tends to push the pipe — and the fireperson holding it — backward, so they must brace hard to keep it steady.

AnswerThe water's reaction force (Newton's 3rd law) pushes the pipe backward, so the fireperson struggles to hold it.
10
A spacecraft is in a region of space where gravity is negligible. Suggest how it can change its velocity.
ICON Solution

It can fire its engine to eject gas (exhaust) in one direction. By Newton's third law, the ejected gas pushes the spacecraft with an equal and opposite force in the other direction.

This thrust changes the spacecraft's velocity — speeding it up, slowing it down, or turning it, depending on the direction the gas is expelled.

AnswerBy expelling exhaust gases from its engine — the reaction force (rocket propulsion) changes its velocity.
3

Exercise

Revise, Reflect, Refine — end-of-chapter questions.

1
Using a horizontal force $F$, a table is moved across the floor at constant velocity. How much is the frictional force exerted by the floor on the table?
ICON Solution

Constant velocity ⇒ zero acceleration ⇒ net force is zero. The only two horizontal forces are the applied force $F$ (forward) and friction $f$ (backward). For them to cancel:

$F - f = 0 \Rightarrow f = F$

AnswerThe frictional force equals $F$ (same magnitude, opposite direction).
2
For a ball on a smooth (frictionless) surface, choose the correct option:
  • If no net force is applied, the velocity will remain the same / increase / decrease.
  • If a net force acts in the direction of motion, the speed will remain the same / increase / decrease.
  • If a net force acts opposite to the motion, the speed will remain the same / increase / decrease.
ICON Solution
  • (i) No net force ⇒ no acceleration ⇒ velocity will remain the same (Newton's first law).
  • (ii) Force along the motion adds to the speed ⇒ speed will increase.
  • (iii) Force opposite to the motion slows it down ⇒ speed will decrease.
Answer(i) remain the same  •  (ii) increase  •  (iii) decrease
3
Forces of 5 N and 4 N act in opposite directions on block P, while block Q moves with a constant velocity (Fig. 6.36). Which statement is correct?
P5 N4 N(a) Block PQconstant v(b) Block Q
Fig. 6.36: Block P with two forces; block Q at constant velocity
  • P experiences a net force and Q does not.
  • P does not experience a net force and Q does.
  • Both P and Q experience a net force.
  • Neither P nor Q experiences a net force.
ICON Solution

Block P: two opposite forces ⇒ net force $= 5 - 4 = 1\ \text{N} \neq 0$ ⇒ P has a net force.

Block Q: moving at constant velocity ⇒ zero acceleration ⇒ net force is zero.

Answer(i) P experiences a net force and Q does not.
4
In a snake-boat race (Vallam kali), 100 oarsmen row. 95 row backward to push the boat forward, but 5 mistakenly row in the opposite direction. Each applies 200 N. Find the net force on the boat. (Ignore drag.)
ICON Solution
  1. Forward force (95 oarsmen)$95 \times 200 = 19000\ \text{N}$
  2. Backward force (5 oarsmen)$5 \times 200 = 1000\ \text{N}$
  3. Net force$19000 - 1000 = 18000\ \text{N}$ (forward)
Answer18000 N in the forward direction.
5
When a net force acts on an object, the object accelerates:
  • opposite to the force, with acceleration proportional to the force.
  • opposite to the force, with acceleration proportional to the mass.
  • in the direction of force, with acceleration inversely proportional to the force.
  • in the direction of force, with acceleration proportional to the force.
ICON Solution

Newton's second law: $a = \dfrac{F}{m}$. The acceleration is in the same direction as the net force and is directly proportional to it (for fixed mass).

Answer(iv) in the direction of force, with acceleration proportional to the force.
6
Position–time graphs for four objects A, B, C, D are shown (Fig. 6.37). A net force acts on which object?
TimePosition0Object ATimePosition0Object BTimePosition0Object CTimePosition0Object D
Fig. 6.37: Position–time graphs for objects A, B, C and D
ICON Solution

A net force exists only when there is acceleration, i.e. when the velocity (the slope of a position–time graph) is changing — a curved graph.

  • A: straight inclined line ⇒ constant velocity ⇒ no net force.
  • B: horizontal line ⇒ at rest ⇒ no net force.
  • C: curved (slope keeps increasing) ⇒ accelerating ⇒ net force acts.
  • D: straight declining line ⇒ constant (negative) velocity ⇒ no net force.
Answer(iii) A net force acts on Object C.
7
A sailor jumps from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why?
A sailor jumping from a boat to the shore
Fig. 6.38: A sailor jumping forward
ICON Solution

To jump forward, the sailor's feet push the boat backward. By Newton's third law, the boat pushes the sailor forward with an equal and opposite force (letting them reach the shore).

The reaction on the boat therefore acts backward, so the boat moves backward — away from the shore, in the direction opposite to the sailor's jump.

AnswerYes — the boat moves backward (opposite to the jump), due to Newton's third law.
8
During a high jump, a landing mat or sand bed is placed for the athlete to fall on (Fig. 6.39). Explain the reason.
A landing mat for a high jump event
Fig. 6.39: A landing mat for a high jump event
ICON Solution

When the athlete lands, they must be brought to rest. A soft mat or sand bed increases the time taken to stop, compared with a hard floor.

A longer stopping time gives a smaller deceleration, and by $F = ma$, a smaller force acts on the athlete's body. This reduces the risk of injury.

AnswerThe mat lengthens the stopping time, lowering the force on the athlete, preventing injury.
9
A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
  • the loaded cart exerts a larger force on the empty cart.
  • the empty cart exerts a larger force on the loaded cart.
  • neither cart exerts a force on the other.
  • both carts exert equal-magnitude forces on each other.
ICON Solution

By Newton's third law, whenever two objects interact, they exert equal and opposite forces on each other — regardless of their masses or loads. (The loaded cart will accelerate less because of its larger mass, but the forces are equal.)

Answer(iv) Both carts exert equal-magnitude forces on each other.
10
The acceleration–mass graph for a fixed force acting on objects of different masses is given (Fig. 6.40). Plot the force–mass graph for this case.
Mass (kg)Acceleration (m s⁻²)0123452.557.510
Fig. 6.40: Given acceleration–mass graph
ICON Solution

Read values from the curve: at $m=1\,$kg, $a=10$; at $m=2\,$kg, $a=5$; at $m=4\,$kg, $a=2.5$. In every case:

$F = ma = 1\times10 = 2\times5 = 4\times2.5 = 10\ \text{N}$

The force is the same (10 N) for all masses. So the force–mass graph is a horizontal straight line at $F = 10\ \text{N}$.

Mass (kg)Force (N)01234551015F = 10 N (constant)
Force–mass graph (the required plot)
AnswerA horizontal line at $F = 10\ \text{N}$ — the force is constant for all masses.
11
The velocity–time graph of an object of mass 10 kg moving in a straight line is shown (Fig. 6.41). Calculate the force acting on the object.
Time (s)Velocity (m s⁻¹)048102030
Fig. 6.41: Velocity–time graph
ICON Solution
  1. Read the graphAt $t=0$, $v=10\ \text{m s}^{-1}$; at $t=8\ \text{s}$, $v=30\ \text{m s}^{-1}$.
  2. Acceleration = slope$a = \dfrac{30-10}{8-0} = \dfrac{20}{8} = 2.5\ \text{m s}^{-2}$
  3. Force$F = ma = 10 \times 2.5 = 25\ \text{N}$
Answer25 N (in the direction of motion).
12
A bullet of mass 50 g moving at 100 m s⁻¹ enters a heavy wooden block and stops after penetrating 50 cm. Estimate the stopping force (assume constant acceleration).
ICON Solution
  1. Convert units$m = 50\ \text{g} = 0.05\ \text{kg}$,   $s = 50\ \text{cm} = 0.5\ \text{m}$,   $u = 100\ \text{m s}^{-1}$,   $v = 0$
  2. Find accelerationUsing $v^2 = u^2 + 2as$:
    $0 = (100)^2 + 2a(0.5) \Rightarrow 0 = 10000 + a \Rightarrow a = -10000\ \text{m s}^{-2}$
  3. Find force$F = ma = 0.05 \times (-10000) = -500\ \text{N}$

The minus sign shows the force opposes the bullet's motion.

AnswerStopping force ≈ 500 N (opposing the motion).
13
A footballer kicks a ball to a speed of 108 km h⁻¹. The estimated force imparted was 800 N and the ball's mass is 0.4 kg. Find the time of contact between foot and ball.
ICON Solution
  1. Convert speed$v = 108\ \text{km h}^{-1} = 108 \times \dfrac{1000}{3600} = 30\ \text{m s}^{-1}$,   $u = 0$
  2. Acceleration$a = \dfrac{F}{m} = \dfrac{800}{0.4} = 2000\ \text{m s}^{-2}$
  3. Time of contactUsing $v = u + at$:   $30 = 0 + 2000\,t \Rightarrow t = \dfrac{30}{2000} = 0.015\ \text{s}$
Answert = 0.015 s (15 milliseconds).
14
An object of mass 2 kg moving at a constant 10 m s⁻¹ enters a rough patch where friction on it is 7 N. At the same moment an extra 3 N force, opposing the motion, is applied. How far does the object travel before stopping?
ICON Solution
  1. Total opposing force$F = 7 + 3 = 10\ \text{N}$ (backward)
  2. Acceleration (deceleration)$a = \dfrac{F}{m} = \dfrac{-10}{2} = -5\ \text{m s}^{-2}$
  3. Distance to stopUsing $v^2 = u^2 + 2as$ with $v=0,\ u=10$:
    $0 = (10)^2 + 2(-5)s \Rightarrow 0 = 100 - 10s \Rightarrow s = 10\ \text{m}$
AnswerThe object travels 10 m before coming to rest.
15
A tractor pulls a harrow of mass $m_1$ with net force $F$, giving acceleration $a_1$. The same tractor pulls a trolley of mass $m_2$ with force $F$, giving $a_2$. If it now pulls the trolley with the harrow placed on it (same force $F$), find the acceleration in terms of $a_1$ and $a_2$. (Ignore friction.)
ICON Solution
  1. Express each mass$F = m_1 a_1 \Rightarrow m_1 = \dfrac{F}{a_1}$,    $F = m_2 a_2 \Rightarrow m_2 = \dfrac{F}{a_2}$
  2. Combined systemTotal mass $= m_1 + m_2$. New acceleration:
    $a = \dfrac{F}{m_1 + m_2} = \dfrac{F}{\frac{F}{a_1} + \frac{F}{a_2}}$
  3. Simplify$a = \dfrac{F}{F\left(\frac{1}{a_1} + \frac{1}{a_2}\right)} = \dfrac{1}{\frac{1}{a_1}+\frac{1}{a_2}} = \dfrac{a_1 a_2}{a_1 + a_2}$
Answer$a = \dfrac{a_1 a_2}{a_1 + a_2}$
16
A bar magnet is brought near a magnetic compass (Fig. 6.42). By Newton's third law the two exert equal and opposite forces, yet the compass needle moves while the bar magnet does not. Explain why.
A bar magnet and a magnetic compass
Fig. 6.42: A bar magnet and a magnetic compass
ICON Solution

The magnetic forces on the two are indeed equal in magnitude. But the resulting acceleration depends on mass, since $a = \dfrac{F}{m}$.

The compass needle is very light, so the same force produces a large acceleration and it swings noticeably. The bar magnet is much heavier, so the equal force gives it only a negligible acceleration — it barely moves.

AnswerEqual forces, but the light needle accelerates strongly while the heavy magnet accelerates negligibly.

That's the full Chapter 6 — solved!

Examples · In-Text “Pause & Ponder” · Exercise “Revise, Reflect, Refine”

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