Vardaan Learning Institute
NCERT SOLUTIONS • CHAPTER 6
How Forces Affect Motion (Revise, Reflect, Refine)
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?
Solution
The frictional force exerted by the floor on the table is equal in magnitude to F and acts in the opposite direction ($-F$).
Reasoning:
- When the table moves with a constant velocity, its acceleration is zero ($a = 0$).
- According to Newton's Second Law ($F_{net} = ma$), the net horizontal force acting on the table must be zero ($F_{net} = 0$).
- Therefore, the applied forward force $F$ and the opposing frictional force $f_{friction}$ balance each other completely: $F - f_{friction} = 0 \implies f_{friction} = F$.
For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct:
(i) If no net force is applied on the ball, the velocity of the ball will remain the same / increase / decrease.
(ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same / increase / decrease.
(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same / increase / decrease.
Solution
(i) If no net force is applied on the ball: The velocity of the ball will remain the same (Newton's First Law of Motion / Inertia).
(ii) If a net force is applied in the direction of motion: The magnitude of the velocity of the ball will increase (positive acceleration).
(iii) If a net force is applied opposite to the direction of motion: The magnitude of the velocity of the ball will decrease (retardation/deceleration).
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 statement is correct?
(i) P experiences a net force and Q does not experience a net force.
(ii) P does not experience a net force and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.
Solution
Correct Option: (i) P experiences a net force and Q does not experience a net force.
Reasoning:
- Block P: Two unequal forces ($5\text{ N}$ right, $4\text{ N}$ left) act on P. Net force on P $= 5\text{ N} - 4\text{ N} = 1\text{ N}$ towards the right $\implies$ P experiences an unbalanced net force.
- Block Q: Block Q is moving with a constant velocity ($\vec{a} = 0$). By Newton's First Law, when velocity is constant, the net force acting on Q is zero ($F_{net} = 0$).
While practising for the snake boat race (Vallum 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.)
Solution
Given:
- Force exerted by 1 oarsman = $200 \text{ N}$
- Number of oarsmen rowing forward-propelling force ($F_{forward}$) = $95$
- Number of oarsmen rowing opposing force ($F_{opposing}$) = $5$
Calculations:
$$F_{forward} = 95 \times 200 \text{ N} = 19,000 \text{ N (forward)}$$
$$F_{opposing} = 5 \times 200 \text{ N} = 1,000 \text{ N (backward)}$$
$$\text{Net Force} = F_{forward} - F_{opposing} = 19,000 \text{ N} - 1,000 \text{ N} = \mathbf{18,000\text{ N (in the forward direction)}}$$
When a net force acts on an object, we observe that the object accelerates:
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.
Solution
Correct Option: (iv) in the direction of force, with acceleration proportional to the force acting on the object.
Reasoning:
According to Newton's Second Law of Motion ($a = \frac{F}{m}$), acceleration is directly proportional to the net force applied ($a \propto F$) and acts in the exact same direction as the net force.
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:
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D
Solution
Correct Option: (iii) Object C
Justification:
- Object A: Straight line with positive constant slope $\implies$ constant velocity $\implies a = 0 \implies F_{net} = 0$.
- Object B: Horizontal line $\implies$ position constant / object at rest $\implies a = 0 \implies F_{net} = 0$.
- Object C: Curved line (parabola) $\implies$ changing slope $\implies$ changing velocity $\implies$ non-zero acceleration ($a \ne 0$) $\implies$ net force acts on Object C.
- Object D: Straight line with negative constant slope $\implies$ constant velocity $\implies a = 0 \implies F_{net} = 0$.
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.
Solution
Yes, the boat will move in the backward direction (away from the shore).
Reasoning (Newton's Third Law of Motion):
- When the sailor jumps forward onto the shore, his feet apply an action force pushing the boat backwards.
- According to Newton's Third Law of Motion (To every action there is an equal and opposite reaction), the boat simultaneously exerts an equal reaction force pushing the sailor forward onto the shore.
- Since the boat floats on water with low friction, the backward action force pushes the boat backwards into the water.
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.
Solution
Scientific Reason (Newton's Second Law & Impulse):
- When an athlete falls from a high jump onto a soft mat or sand bed, the mat depresses, increasing the time interval ($\Delta t$) required for the athlete's high downward velocity to reduce to zero.
- From $F = \frac{\Delta p}{\Delta t}$, increasing the stopping time duration significantly decreases the magnitude of the impact force ($F$) exerted on the athlete's body.
- If the athlete landed on a hard floor, $\Delta t$ would be near zero, generating a massive impact force that could cause bone fractures or severe injuries.
A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
(i) the loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
Solution
Correct Option: (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
Reasoning:
By Newton's Third Law of Motion, action and reaction forces between two colliding bodies are always equal in magnitude and opposite in direction, regardless of their masses, speeds, or load states.
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.
Solution
1. Calculating Constant Force ($F$):
From Fig. 6.40, reading mass ($m$) and acceleration ($a$) data points:
- At $m = 1 \text{ kg}$, $a = 10 \text{ m/s}^2 \implies F = m \cdot a = 1 \times 10 = 10 \text{ N}$
- At $m = 2 \text{ kg}$, $a = 5 \text{ m/s}^2 \implies F = 2 \times 5 = 10 \text{ N}$
- At $m = 4 \text{ kg}$, $a = 2.5 \text{ m/s}^2 \implies F = 4 \times 2.5 = 10 \text{ N}$
2. Force-Mass Graph Plot:
Since the applied force is constant at $F = 10\text{ N}$ for all masses, the Force-Mass graph is a horizontal straight line parallel to the Mass axis (X-axis) at $F = 10\text{ N}$.
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.
Solution
From Fig. 6.41:
- Mass of object, $m = 10 \text{ kg}$
- At $t = 0 \text{ s}$, initial velocity $u = 10 \text{ m/s}$
- At $t = 8 \text{ s}$, final velocity $v = 30 \text{ m/s}$
1. Acceleration ($a$):
$$a = \frac{\Delta v}{\Delta t} = \frac{30 - 10}{8 - 0} = \frac{20}{8} = 2.5 \text{ m/s}^2$$
2. Net Force ($F$):
$$F = m \cdot a = 10 \text{ kg} \times 2.5 \text{ m/s}^2 = \mathbf{25\text{ N}}$$
A bullet of mass 50 g moving with a speed of 100 m/s enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).
Solution
Given:
- Mass of bullet, $m = 50 \text{ g} = 0.05 \text{ kg}$
- Initial velocity, $u = 100 \text{ m/s}$
- Final velocity, $v = 0 \text{ m/s}$
- Distance, $s = 50 \text{ cm} = 0.5 \text{ m}$
1. Acceleration ($a$):
$$v^2 = u^2 + 2as \implies 0^2 = (100)^2 + 2 a (0.5) \implies 0 = 10000 + 1 a \implies a = -10,000 \text{ m/s}^2$$
2. Stopping Force ($F$):
$$F = m \cdot a = 0.05 \text{ kg} \times (-10,000 \text{ m/s}^2) = \mathbf{-500\text{ N}}$$
The retarding stopping force exerted by the wooden block is $500\text{ N}$ in the direction opposite to motion.
An ace footballer converted a penalty shot by kicking the football with a speed of 108 km/h. 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.
Solution
Given:
- Initial velocity before kick, $u = 0 \text{ m/s}$
- Final velocity after kick, $v = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$
- Force, $F = 800 \text{ N}$
- Mass of football, $m = 0.4 \text{ kg}$
Using Impulse-Momentum Theorem ($F \cdot \Delta t = m \Delta v$):
$$800 \times \Delta t = 0.4 \times (30 - 0) = 12 \text{ N}\cdot\text{s}$$
$$\Delta t = \frac{12}{800} = 0.015 \text{ s} = \mathbf{15\text{ milliseconds}}$$
An object of mass 2 kg moving with a constant velocity of 10 m/s 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?
Solution
Given:
- Mass, $m = 2 \text{ kg}$
- Initial velocity, $u = 10 \text{ m/s}$
- Final velocity, $v = 0 \text{ m/s}$
- Frictional opposing force, $f_k = 7 \text{ N}$
- Additional opposing force, $F_{ext} = 3 \text{ N}$
1. Net Opposing Force ($F_{net}$):
$$F_{net} = 7 \text{ N} + 3 \text{ N} = 10 \text{ N (opposing)}$$
2. Deceleration ($a$):
$$a = -\frac{F_{net}}{m} = -\frac{10}{2} = -5 \text{ m/s}^2$$
3. Stopping Distance ($s$):
$$v^2 = u^2 + 2as \implies 0^2 = 10^2 + 2(-5)s \implies 0 = 100 - 10s \implies 10s = 100 \implies s = \mathbf{10\text{ m}}$$
A tractor pulls a harrow (a ploughing tool) of mass $m_1$ with a net force F resulting in an acceleration of $a_1$. The same tractor pulls a trolley of mass $m_2$ with a force F producing an acceleration of $a_2$. 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 $a_1$ and $a_2$. Ignore friction.
Solution
1. Individual Mass Expressions:
$$F = m_1 a_1 \implies m_1 = \frac{F}{a_1}$$
$$F = m_2 a_2 \implies m_2 = \frac{F}{a_2}$$
2. Combined System Acceleration ($a_{comb}$):
$$\text{Total Mass } M = m_1 + m_2 = \frac{F}{a_1} + \frac{F}{a_2} = F \left(\frac{1}{a_1} + \frac{1}{a_2}\right) = F \left(\frac{a_1 + a_2}{a_1 a_2}\right)$$
$$a_{comb} = \frac{F}{M} = \frac{F}{F \left(\frac{a_1 + a_2}{a_1 a_2}\right)} = \mathbf{\frac{a_1 a_2}{a_1 + a_2}}$$
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.
Solution
Scientific Explanation:
- According to Newton's Third Law, the magnetic force exerted by the bar magnet on the compass needle ($F_{NC}$) is equal in magnitude and opposite in direction to the magnetic force exerted by the needle on the bar magnet ($F_{CN}$).
- However, acceleration depends inversely on mass ($a = \frac{F}{m}$):
- The compass needle has an extremely small mass ($m_{needle}$), so equal force $F$ produces a large, highly noticeable angular acceleration ($a_{needle} = \frac{F}{m_{needle}}$).
- The bar magnet (and the table/hand holding it) has a vastly larger mass ($M_{magnet} \gg m_{needle}$), making its acceleration ($a_{magnet} = \frac{F}{M_{magnet}}$) imperceptibly small and virtually zero.
