๐Ÿ’ฌ Think about it

Push a ball gently and it crawls; push it hard and it shoots off. A force produces acceleration — but how much? Does it depend on the object's mass too? Newton's second law answers this precisely. Let us build up to it.

What does a net force produce?
  • A force can start, stop or change the velocity of an object.
  • A change in velocity means the object is accelerating.
  • So a force produces acceleration.
Example: A gentle push gives a ball a small acceleration; a strong push gives a large one.
๐Ÿ”ง Activity 6.3 — Let us experiment (Demonstration activity)

In this Activity, we will pull a cart with different forces (using different falling weights) to see how the acceleration changes for a fixed mass.

Procedure
1. Make a cart from a cardboard box with pencil axles and four ball-bearing wheels, with a thread to pull it.
2. Run the thread over a small pipe (pulley) at the table edge and attach a cup to hold weights. The falling cup pulls the cart with a constant force (Fig. 6.17).
3. Measure the mass of the cup and its contents.
4. Record the cart in slow motion as it travels a fixed distance, and find the time T 1 .
5. Double the mass in the cup (which doubles the pulling force) and repeat to get time T 2 .
Analysis
Both runs start from rest (u = 0) and cover the same distance s. Using \( s = \dfrac{1}{2}aT^2 \) for each run and equating gives \( \dfrac{a_1}{a_2} = \dfrac{T_2^{\,2}}{T_1^{\,2}} \). The values show that increasing the force (for the same mass) increased the acceleration.
โ—† Summary
  • Pull a cart with a falling cup.
  • Time it over a fixed distance.
  • Double the force and retime.
  • More force = more acceleration.
๐Ÿ”ง Activity 6.4 — Let us experiment (Demonstration activity)

In this Activity, we will keep the force fixed but change the cart's mass to see how mass affects the acceleration.

Procedure
1. Repeat Activity 6.3, but keep the cup and its contents (the force) constant. Double the mass of the cart by adding objects to it.
2. Measure the cart's mass with its contents.
3. Carry out the timing steps of Activity 6.3.
Analysis
Find the ratio of accelerations for the two cases. For the same force, doubling the cart's mass decreased its acceleration. So acceleration is inversely related to mass.
โ—† Summary
  • Keep the pulling force fixed.
  • Double the cart's mass.
  • Time it again.
  • More mass = less acceleration.
What is Newton's second law of motion?
  • When a net force acts on an object, it accelerates in the direction of that force.
  • The acceleration is proportional to the net force.
  • The acceleration is inversely proportional to the mass of the object.
\( a \)
=
\( \dfrac{F}{m} \)  or  \( F = ma \)
How is one newton defined?
  • Using F = ma with m = 1 kg and a = 1 m s −2 , F = 1 kg m s −2 = 1 N.
  • One newton is the force that gives a 1 kg object an acceleration of 1 m s −2 .
\( 1\ \text{N} \)
=
\( 1\ \text{kg} \times 1\ \text{m s}^{-2} = 1\ \text{kg m s}^{-2} \)
Example: Holding a 100 g mass in your palm, the upward force you apply is about 1 N.
What is the gravitational force on an object (F = mg)?
  • An object falls towards the Earth with acceleration due to gravity, g.
  • The gravitational force on a mass m is F = mg.
  • Near the Earth's surface g = 9.8 m s −2 (about 10 m s −2 for quick estimates).
\( F \)
=
\( mg \)
๐Ÿ“ Note
  • The acceleration due to the Earth's gravitational force (g) does not depend on the mass of the object.
โœŽ Example 6.4

A barbell has 10 kg on each side of a 10 kg bar (Fig. 6.8). How much force keeps it steady?

Total mass = 10 + 10 + 10 = 30 kg. Gravitational force, \( F = mg = 30 \times 9.8 = 294\ \text{N} \) downward.

To hold it steady, the lifter applies an equal force upward = 294 N, upward .

โœŽ Example 6.5

A 25 kg block faces a maximum friction of 50 N. Find its displacement in 2 s when pushed with (i) 50 N and (ii) 55 N.

(i) Applied force = friction = 50 N, so net force = 0. The block stays stationary .

(ii) Net force = 55 − 50 = 5 N. Acceleration \( a = \dfrac{F}{m} = \dfrac{5}{25} = 0.2\ \text{m s}^{-2} \).

Displacement \( s = ut + \dfrac{1}{2}at^2 = 0 + \dfrac{1}{2}\times0.2\times(2)^2 = 0.4\ \text{m} \) in the forward direction.

โœŽ Example 6.6

A 1500 kg sports car moves east; its velocity-time graph is shown in Fig. 6.21. Find the force during (i) 0–5 s, (ii) 5–10 s, (iii) 10–15 s.

(i) u = 0, v = 10 m s −1 , t = 5 s. Using v = u + at, \( a = \dfrac{10}{5} = 2\ \text{m s}^{-2} \). Force \( F = ma = 1500\times2 = 3000\ \text{N} \), east.

(ii) The graph is flat (constant velocity), so acceleration is zero and no force acts.

(iii) u = 10, v = 0, t = 5 s, so \( a = \dfrac{0-10}{5} = -2\ \text{m s}^{-2} \). Force \( F = 1500\times(-2) = -3000\ \text{N} \) — that is 3000 N towards the west (opposite to motion).

Important Points
  • Newton's second law: F = ma; acceleration is along the net force.
  • One newton accelerates 1 kg at 1 m s −2 .
  • Weight is the gravitational force, F = mg, with g = 9.8 m s −2 .
๐Ÿงฉ Threads of Curiosity
  • In Activity 6.3, doubling the force should double the acceleration, but the increase is often a little less.
  • In Activity 6.4, doubling the mass should halve the acceleration, but the value may differ slightly.
  • Apart from measurement errors, friction between the wheels and the surface causes these differences.
๐Ÿ“š Ready to Go Beyond
  • The more complete form of Newton's second law uses momentum — the product of mass and velocity.
  • It states that the rate of change of momentum is proportional to the net force, in its direction.
  • This form works even when the object's mass is not constant.
๐ŸŒŽ Bridging Science and Society
  • A fielder pulls their hands back while catching a fast ball, increasing the stopping time so the force (and injury) is smaller.
  • Airbags inflate into a soft cushion in a crash, increasing the time over which a passenger stops, reducing the force on them.
  • Cracking a coconut works the opposite way — it stops in a very short time, so the ground exerts a very large force that breaks the shell.
โ“ Test Yourself
  1. Write Newton's second law as a formula.
    View Answer Hide Answer
    F = ma (or a = F/m).
  2. A 2 kg object accelerates at 3 m s−². What net force acts?
    View Answer Hide Answer
    F = ma = 2 × 3 = 6 N.
  3. What is the value of g near the Earth's surface?
    View Answer Hide Answer
    9.8 m s−² (about 10 m s−²).
Important Definitions
  • Newton's second law of motion — net force equals mass times acceleration (F = ma); acceleration is along the net force.
  • Acceleration due to gravity (g) — the acceleration of a freely falling object due to the Earth's gravitational force, 9.8 m s −2 .

๐Ÿ“‹ NCERT Question 5 — When a net force

When a net force acts on an object, choose how it accelerates relative to the force.
View Answer →

๐Ÿ“‹ NCERT Question 8 — During a high jump

Why is a soft landing mat or sand bed placed for a high-jump athlete to fall on?
View Answer →

๐Ÿ“‹ NCERT Question 11 — The velocity-time graph of

From the velocity-time graph (Fig. 6.41), calculate the force on a 10 kg object.
View Answer →

๐Ÿ“‹ NCERT Question 12 — A bullet of mass

A 50 g bullet at 100 m s−¹ stops in 50 cm — estimate the stopping force.
View Answer →

๐Ÿ“‹ NCERT Question 13 — An ace footballer converted

A football kicked at 108 km/h with 800 N force, mass 0.4 kg — find the contact time.
View Answer →

๐Ÿ“‹ NCERT Question 14 — An object of mass

A 2 kg object at 10 m s−¹ meets friction 7 N plus 3 N opposing force — find the distance before stopping.
View Answer →

๐Ÿ“‹ NCERT Question 10 — The acceleration-mass graph for

From the acceleration-mass graph (Fig. 6.40), plot the force-mass graph.
View Answer →
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