These questions appear inside the chapter itself, not at the end. The book page number is shown against each one so you can find it while reading.
A weightlifter lifts a barbell (Fig. 6.8). List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?
The two forces acting on the barbell are:
Are they balanced? Yes. When the weightlifter keeps the barbell steady, the barbell is at rest — its velocity is zero and unchanging, so its acceleration is zero. By Newton’s first law, a body with zero acceleration has zero net force on it. Therefore the upward force of the weightlifter must exactly equal the downward weight of the barbell: the two forces are equal in magnitude and opposite in direction, i.e. they are balanced.
Two players R and S are participating in an arm-wrestling match (Fig. 6.9). At the instant when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?
No, at that instant the forces are not balanced.
In arm-wrestling each player pushes against the other. As long as the joined hands stay still in the middle, the two pushes are equal and opposite — they are balanced and there is no motion. But the moment the arms begin to tilt (move) to the front, the joined hands have started to accelerate in that direction. A change in motion can only be produced by a net (unbalanced) force, so at that instant the two forces are no longer equal.
Which player exerted the larger force? The hands always move towards the side of the weaker push, i.e. in the direction of the stronger player’s force. So the player who is pushing the joined hands in the direction they are actually tilting (out of the page, towards you) is exerting the larger force — that player is winning the match at that instant.
An object is moving with a constant velocity. Is there a net force acting upon it?
No, there is no net force acting on it.
‘Constant velocity’ means that both the speed and the direction of the object are unchanging. If the velocity is not changing, the object’s acceleration is zero. By Newton’s second law, net force = mass × acceleration, so a zero acceleration means the net force is zero.
Several individual forces may still be acting on the object (for example friction, gravity, a normal force or an applied push), but they must all cancel out to give a zero net force. This is exactly what Newton’s first law tells us: an object keeps moving with a constant velocity unless a net force acts on it.
Suppose, no net force is acting on an object. Which of the following situations are possible?
If no net force acts on an object, its acceleration must be zero (since net force = mass × acceleration). Checking each situation:
So situations (i) and (ii) are possible, while (iii) is not possible.
In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.
In everyday life, several forces usually act on an object at the same time — gravity, the normal force from a surface, friction, an applied push, and so on — so a truly force-free object is very rare. However, we can arrange these forces so that they cancel one another, giving a zero net force even though many forces are present.
Example 1 — a book resting on a table. Gravity pulls the book downwards, and the table pushes it upwards with an equal normal force. These two balance, so the net force is zero and the book stays at rest.
Example 2 — pushing a box at a steady speed. When you push a heavy box across the floor so that it moves at a constant velocity, the friction opposing its motion is exactly matched by the force you apply. The applied force and friction cancel, the net force is zero, and the box keeps moving at constant velocity.
In each case an additional force (the normal force, or your applied push) is deliberately balanced against the other forces so that the total, or net, force works out to zero.
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 the toy car?
The net force on the toy car is zero.
The toy car moves with a constant velocity, so its velocity is not changing and its acceleration is zero. By Newton’s second law of motion,
net force = mass × acceleration = 0.1 kg × 0 m s−2 = 0 N.
(The mass, 100 g = 0.1 kg, and the speed, 0.5 m s−1, are given only to check whether you realise that constant velocity always means zero acceleration and hence zero net force — no matter what the mass or the speed happens to be.)
Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.
You would need to apply the larger force to the heavier child.
By Newton’s second law, force = mass × acceleration (F = ma). To give both children the same acceleration a, the force needed is directly proportional to the child’s mass: the greater the mass, the greater the force required.
The heavier child has more mass, so for the same initial acceleration a larger force must be applied. For example, if one child has twice the mass of the other, you would have to push twice as hard on the heavier child to produce the same acceleration.
How are glass items packed for transportation using a bubble wrap or hay protected from damage?
Bubble wrap and hay protect fragile glass by increasing the time over which the glass is brought to a stop during a jerk, bump or collision.
When a packed box is jolted or dropped, the glass inside has to change its velocity (slow down or stop). The force felt by the glass depends on how quickly that change happens: if the glass is stopped almost instantly (a very short time), a very large force acts on it, and it can crack or shatter.
The soft, springy bubble wrap or hay cushions the glass and lets it come to rest gradually, spreading the change of velocity over a longer time. A longer stopping time means a smaller deceleration, and therefore a much smaller force on the glass — so it is far less likely to break. (This is the same principle by which a cricketer draws the hands back while catching a fast ball, and by which airbags protect passengers.)
Why does a fireperson sometimes struggle when holding the pipe issuing water?
This happens because of Newton’s third law of motion — every action has an equal and opposite reaction.
The hose pushes a large amount of water forwards at high speed. By Newton’s third law, the water pushes back on the hose (and on the fireperson holding it) with an equal and opposite force — a strong backward ‘recoil’ or thrust.
Because a large mass of water is being thrown out very quickly, this backward reaction force can be very large. The fireperson therefore has to brace themselves and grip the pipe firmly to keep it steady — the faster and greater the flow of water forward, the harder the pipe pushes back, which is why it can be a real struggle to hold.
Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how can it change its velocity.
With gravity negligible and practically no other external force acting on it in empty space, the spacecraft can change its velocity only by applying a force to itself, using Newton’s third law of motion.
The spacecraft’s engines burn fuel and eject exhaust gases at high speed in one direction. By Newton’s third law, these gases push back on the spacecraft with an equal and opposite force (the thrust) in the other direction. This thrust is the net force that accelerates the spacecraft, changing its speed and/or direction.
So, by firing its rocket engines and expelling gas backwards, the spacecraft experiences a forward reaction force and changes its velocity — the same principle by which a rocket lifts off. To speed up, it ejects gas opposite to its motion; to slow down or turn, it fires the engine in the appropriate direction (this is how the Vikram lander of Chandrayaan-3 slowed itself for a soft landing).
Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
The frictional force is equal to F (equal in magnitude to the applied force), acting in the direction opposite to the motion.
Since the table moves at a constant velocity, its acceleration is zero, so the net force on it is zero (Newton’s first law). Only two horizontal forces act on the table:
For the net horizontal force to be zero, these two must balance exactly. Therefore the frictional force exerted by the floor on the table is equal to F, directed opposite to the applied force.
For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.
Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity. Which of the following statements is correct?
Correct option: (i) — P experiences a net force and Q does not.
So P has a net force while Q does not — option (i) is correct.
While practising for the snake boat race (Vallam kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)
Net force on the snake boat = 18 000 N (18 kN), directed forward.
Group the oarsmen by the direction of the force they produce on the boat:
These two groups push in opposite directions, so the net force is the difference between them:
net force = 19 000 N − 1 000 N = 18 000 N, acting forward (in the direction the 95 oarsmen are propelling the boat).
When a net force acts on an object, we observe that the object accelerates:
Correct option: (iv) — in the direction of the force, with acceleration proportional to the force acting on the object.
Newton’s second law of motion says that acceleration a = F ÷ m. From this:
Hence the correct choice is (iv).
The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
Correct option: (iii) — a net force acts on Object C.
A net force produces an acceleration, i.e. a changing velocity. On a position-time graph, a changing velocity shows up as a curved line, while a straight line (including a horizontal one) means the velocity is constant and hence the net force is zero.
Only Object C’s graph is curved, so a net force acts only on Object C — option (iii).
A sailor jumps out 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?
Yes, the boat will move — it moves backward, in the direction opposite to the sailor’s jump.
This is a direct example of Newton’s third law of motion. To jump forward towards the shore, the sailor pushes backward on the boat with their feet. By Newton’s third law, the boat pushes the sailor forward with an equal and opposite force, and this is what propels the sailor onto the shore.
At the same time, the sailor’s backward push acts on the boat. Since water offers very little resistance, this force pushes the boat backward, away from the shore. So, as the sailor jumps forward, the boat recoils in the opposite (backward) direction. (This is why one must step carefully out of a small boat.)
During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.
The soft mat or sand bed protects the athlete by increasing the time taken to stop when they land.
When the athlete falls, they are moving fast and must be brought to rest. If they landed on a hard surface, they would stop almost instantly — a very short stopping time. Because the force depends on how quickly the velocity changes, a sudden stop means a very large force acts on the body, which can cause serious injury.
A soft mat or sand bed compresses and lets the athlete sink in and come to rest gradually, spreading the change of velocity over a longer time. A longer stopping time means a smaller deceleration, and therefore a much smaller force on the body — greatly reducing the risk of injury. (This is the same idea used in airbags, bubble wrap, and drawing back the hands while catching a cricket ball.)
A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
Correct option: (iv) — both carts exert an equal magnitude of force on each other.
By Newton’s third law of motion, whenever two objects interact they exert equal and opposite forces on each other — always, regardless of their masses or how they are moving. So during the collision, the loaded cart and the empty cart push on each other with forces that are equal in magnitude and opposite in direction.
(Their accelerations will differ — the lighter empty cart is pushed back more strongly because it has less mass — but the forces they exert on one another are equal.) Hence option (iv) is correct.
The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.
Reading the acceleration-mass graph. As the mass increases, the acceleration decreases in such a way that the product mass × acceleration stays the same. Reading a few points from Fig. 6.40 and using F = ma:
| Mass m (kg) | Acceleration a (m s−2) | Force F = ma (N) |
|---|---|---|
| 1 | 10 | 10 |
| 2 | 5 | 10 |
| 4 | 2.5 | 10 |
In every case the force works out to the same value, F = 10 N. This makes sense: the same force was applied to each object, so the acceleration is inversely proportional to the mass (a ∝ 1/m), which is why the acceleration-mass graph is a falling curve (a rectangular hyperbola).
The force-mass graph is therefore a horizontal straight line at F = 10 N, parallel to the mass axis — the force does not change as the mass changes:
The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
Step 1 — read the graph. From Fig. 6.41 the straight line gives:
Step 2 — find the acceleration (the slope of the line). Since the graph is a straight line, the acceleration is constant:
a = (v − u) ÷ t = (30 − 10) ÷ (8 − 0) = 20 ÷ 8 = 2.5 m s−2.
Step 3 — apply Newton’s second law with m = 10 kg:
F = ma = 10 kg × 2.5 m s−2 = 25 N.
So the force acting on the object is 25 N, in the direction of motion (the object is speeding up).
A bullet of mass 50 g moving with a speed of 100 m s−1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume the bullet undergoes constant acceleration within the block).
Given: mass m = 50 g = 0.05 kg; initial speed u = 100 m s−1; final speed v = 0 (it stops); distance travelled s = 50 cm = 0.5 m.
Step 1 — find the (constant) acceleration using v2 = u2 + 2as:
0 = (100)2 + 2 × a × 0.5
0 = 10000 + a ⇒ a = −10000 m s−2.
The negative sign shows it is a retardation (the bullet is slowing down).
Step 2 — find the stopping force using Newton’s second law:
F = ma = 0.05 kg × (−10000 m s−2) = −500 N.
So the stopping force has a magnitude of 500 N, acting opposite to the bullet’s motion (that is why it comes out negative). The wooden block exerts a resisting force of about 500 N on the bullet.
An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h−1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.
Step 1 — convert the speed to SI units. The ball starts from rest (u = 0) and is kicked to a speed
v = 108 km h−1 = 108 × (1000 ÷ 3600) m s−1 = 108 × (5 ÷ 18) = 30 m s−1.
Step 2 — find the acceleration from Newton’s second law:
a = F ÷ m = 800 N ÷ 0.4 kg = 2000 m s−2.
Step 3 — find the contact time using v = u + at:
30 = 0 + 2000 × t ⇒ t = 30 ÷ 2000 = 0.015 s.
So the foot was in contact with the ball for 0.015 s (that is, 15 milliseconds).
An object of mass 2 kg moving with a constant velocity of 10 m s−1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?
Step 1 — find the total force opposing the motion. Inside the rough patch two forces act against the object’s motion:
total opposing force = friction + additional force = 7 N + 3 N = 10 N.
Step 2 — find the deceleration using Newton’s second law (m = 2 kg):
a = F ÷ m = 10 N ÷ 2 kg = 5 m s−2, directed opposite to the motion, so a = −5 m s−2.
Step 3 — find the distance using v2 = u2 + 2as, with u = 10 m s−1 and v = 0:
0 = (10)2 + 2 × (−5) × s
0 = 100 − 10s ⇒ s = 100 ÷ 10 = 10 m.
So the object travels 10 m in the rough patch before coming to rest.
A tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.
Step 1 — write each mass in terms of the force and its acceleration using F = ma:
Step 2 — pull both together. When the trolley carries the harrow, the total mass is (m1 + m2), pulled by the same force F. The new acceleration a is:
a = F ÷ (m1 + m2) = F ÷ ( F/a1 + F/a2 ).
Step 3 — simplify. Take F common in the denominator; it cancels with the F on top:
a = F ÷ [ F ( 1/a1 + 1/a2 ) ] = 1 ÷ ( 1/a1 + 1/a2 ).
Combining the two fractions gives the required expression:
a = (a1 a2) ÷ (a1 + a2).
When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
The two forces are indeed equal in magnitude and opposite in direction (Newton’s third law), but equal forces do not produce equal motion. The motion (acceleration) each object gets depends on its mass, through a = F ÷ m — the same force gives a small mass a large acceleration and a large mass a very small acceleration.
So the light needle moves while the massive bar magnet appears to stay still, even though the forces on the two are equal. (This is the same idea as the Earth and a falling fruit, or a gun and a bullet: equal forces, but very different accelerations because of very different masses.)
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