While reading the statement of Newton’s second law, you need to remember these 3 equations:

**Equation 1:**[F_{net}= ma]

**Equation 2:**[p= mv]

**Equation 3:**[F_{net}= m(v-u) / t]

Before moving onto the numericals,

Let’s first understand all the three Newton’s second law equations.

(One by One)

Where,

F_{net} = net force applied on the body, **N**

m = mass of the object, **kg**

a = acceleration of the object, **m/s ^{2}**

Where,

p = Momentum of the object, **kg m/s**

v = velocity of the object, **m/s**

Where,

m (v – u) / t = rate of change in momentum, **kg m/s ^{2}**

mu = initial momentum of the object, **kg m/s**

mv = final momentum of the object, **kg m/s**

u = initial velocity of the object, **m/s**

v = final velocity of the object, **m/s**

t = time, **seconds**

By remembering these three equations in Newton’s 2^{nd} law of motion, you can **easily** solve the problems mentioned below.

(You can practice these problems, Right Now)

## Newton’s Second Law Equation

(F = ma)

**Numerical 1:**

The ball has a mass of 4 kg. How much net force is required to accelerate this ball at a rate of 9 m/s^{2}?

**Solution:****Given data:**

Mass of the ball, m = 4 kg

The net force required, F_{net} =?

Acceleration of the ball, a = 9 m/s^{2}**According to Newton’s 2 ^{nd} law formula,**

F

_{net}= ma

F

_{net}= 4 × 9

F

_{net}= 36 N

Therefore, a net force of

**36 N**is required to accelerate the ball at a rate of 9 m/s

^{2}.

**Numerical 2:**If the object is accelerating forward at a rate of 10 m/s

^{2}, a net force of 15 N acts on it. Calculate the mass of the object.

**Solution:****Given data:**

Acceleration of the object, a = 10 m/s^{2}

Force applied on the object, F_{net} = 15 N

Mass of the object, m =?**According to Newton’s second law formula,**

F_{net} = ma

m = F_{net }/ a

m = 15 / 10

m = 1.5 kg

Therefore, the mass of the object is **1.5 kg.**

**Numerical 3:**

Consider one bicycle which is accelerating at a rate of 3 m/s^{2}. If the mass of the bicycle is 2.5 kg, then how much force is applied on the bicycle?

**Solution:**

**Given data:**

Acceleration of the bicycle, a = 3 m/s^{2}

Mass of the bicycle, m = 2.5 kg

The net force applied on the bicycle, F_{net} =?**According to Newton’s second law formula,**

F_{net} = ma

F_{net} = 2.5 × 3

F_{net} = 7.5 N

Therefore, when the bicycle accelerates at a rate of 3 m/s^{2}, a force of **7.5 N** is applied to it.

**Numerical 4:**

A 20 kg box accelerates at a rate of ‘a’ m/s^{2}. If a 600 N force is applied on the box, at what rate the box will accelerate further?

**Solution:** **Given data:**

Mass of the box, m = 20 kg

The net force applied on the box, F

_{net}= 600 N

Acceleration of the box, a =?

**According to the formula of Newton’s second law,**

F

_{net}= ma

a = F

_{net}/ m

a = 600 / 20

a = 30 m/s

^{2}

Therefore, the box accelerates at a rate of

**30 m/s**when a force of 600 N is applied on it.

^{2}**Numerical 5:**

When a 92 N force is applied on an object, it accelerates further at a rate of 4 m/s^{2}. What is the mass of an object?

**Solution:****Given data:**

The net force applied on the object, F_{net} = 92 N

Acceleration of the object, a = 4 m/s^{2}

Mass of the object, m =?**According to Newton’s second law,**

F_{net} = ma

m = F_{net }/ a

m = 92 / 4

m = 23 kg

Therefore, the mass of the object is **23 kg.**

## Newton’s Second Law Equation

(p = mv)

**Numerical 1:**

A roller having a mass of 2 kg is rolling forward with a velocity of 4 m/s. Calculate its momentum.

**Solution:** **Given data:**

Mass of the roller, m = 2 kg

Velocity of the roller, v = 4 m/s

Momentum of the roller, p =?**According to the momentum formula,**

Momentum (p) = mass (m) × velocity (v)

p = m × v

p = 2 × 4

p = 8 kg m/s

Therefore, the momentum of the roller is **8 kg m/s.**

**Numerical 2:**

When a 6 kg shopping cart is moving with the velocity of ‘v’ m/s, it gains the momentum of 12 kg m/s. Calculate the velocity of the shopping cart.

**Solution:** **Given data:**

Mass of the shopping cart, m = 6 kg

Momentum of the shopping cart, p = 12 kg m/s

Velocity of the shopping cart, v =?**According to the formula of momentum,**

Momentum (p) = mass (m) × velocity (v)

p = m × v

v = p / m

v = 12 / 6

v = 2 m/s

Therefore, the velocity of the shopping cart is **2 m/s.**

**Numerical 3:**

When a dish is thrown from a tall building, it falls with a velocity of 25 m/s. If the dish has a mass of 200 grams, with what momentum it will strike the ground?

**Solution:** **Given data:**

Velocity of the dish, v = 25 m/s

Mass of the dish, m = 200 grams

Now, 1 kg = 1000 grams (So, 200 grams = 0.2 kg)

Momentum of the dish, p =?**According to the formula of momentum,**

Momentum (p) = mass (m) × velocity (v)

p = m × v

p = 0.2 × 25

p = 5 kg m/s

Therefore, a dish strikes the ground with a momentum of **5 kg m/s.**

**Numerical 4:**

A heavy stone of mass 10 kg is thrown from the hill. The stone gains a momentum of 40 kg m/s. With what velocity the stone strikes the ground?

**Solution:** **Given data:**

Mass of the stone, m = 10 kg

Momentum of the stone, p = 40 kg m/s

The velocity of the stone, v =?**According to the formula of momentum,**

Momentum (p) = mass (m) × velocity (v)

p = m × v

v = p / m

v = 40 / 10

v = 4 m/s

Therefore, the stone strikes the ground with a velocity of **4 m/s.**

**Numerical 5:**

A bike running with a velocity of 15 m/s gains a momentum of 1230 kg m/s. Calculate the mass of the bike.

**Solution:** **Given data:**

Velocity of the bike, v = 15 m/s

Momentum of the bike, p = 1230 kg m/s

Mass of the bike, m =?**According to the formula of momentum,**

Momentum (p) = mass (m) × velocity (v)

p = m × v

m = p / v

m = 1230 / 15

m = 82 kg

Therefore, the mass of the bike is **82** **kg.**

## Newton’s Second Law Equation

[F = m(v-u)/t]

**Numerical 1:**

A robot that has a mass of 5 kg starts moving from the rest. The velocity of the robot further increases and reaches up to ‘v’ m/s, when a force of 20 N is applied to it. What will be the speed of the robot after 20 seconds?

**Solution:** **Given data:**

Mass of the robot, m = 5 kg

Force applied on the robot, F_{net} = 20 N

After time, t = 20 seconds

Final velocity of the robot, v =?

Here, the robot is initially **at rest.**

Therefore, the initial velocity of the robot is u = 0 m/s**According to Newton’s second law equation,**

F_{net} = m (v – u) / t

20 = 5 (v – 0) / 20

5 (v – 0) = 20 × 20

5 v = 400

v = 400 / 5

v = 80 m/s

Therefore, after 20 seconds the robot reaches the speed of **80 m/s**.

**Numerical 2:**

A skateboard having a mass of 300 grams is moving forward with the initial velocity of 110 m/s. When a skateboard strikes with the obstacle, an **opposing force** of 5 N acts on it. The speed of the skateboard decreases and reaches 10 m/s. After how much time, the skateboard will come into rest position?

**Solution:** **Given data:**

Mass of the skateboard, m = 300 grams

Now, 1 kg = 1000 grams (So, 300 grams = 0.3 kg)

Initial velocity of the skateboard, u = 110 m/s

Here, the **opposing force** of 5 N is acting on the skateboard.

Force acting on the skateboard, F_{net} = **– 5** N

Final velocity of the skateboard, v = 10 m/s

Time, t =?**As per the Newton’s second law,**

F_{net} = m (v-u) / t

(- 5) = 0.3 (10 – 110) / t

0.3 (10 – 110) = (- 5) × t

0.3 (- 100) = (- 5) × t

(- 30) = (- 5) × t

5 × t = 30

t = 30 / 5

t = 6 seconds

Therefore, after **6 seconds **the skateboard will come into the rest position.

**Numerical 3:**

A ball is moving forward with the initial velocity of 35 m/s. When an **opposing force** of 15 N is applied by the bat, its velocity starts decreasing. If the mass of the ball is 400 grams, what will be the velocity of the ball after 12 seconds?

**Solution:** **Given data:**

Initial velocity of the ball, u = 35 m/s

Here, the opposing force of 15 N is applied to the ball.

The net force applied on the ball, F_{net} = **– 15 N**

Mass of the ball, m = 400 grams

Now, 1 kg = 1000 grams (So, 400 grams = 0.4 kg)

Final velocity of the ball, v =?

After time, t = 12 seconds**According to Newton’s 2 ^{nd} law equation,**

F

_{net}= m (v – u) / t

(- 15) = 0.4 (v – 35) / 12

0.4 (v – 35) = (- 15) × 12

0.4 (v – 35) = (- 180)

v – 35 = (- 180) / (0.4)

v – 35 = 450

v = 450 + 35

v = 485 m/s

Therefore, the velocity of the ball will be

**485 m/s**after 12 seconds.

**Numerical 4:**

When a force of 7 N is applied on a rubber tyre, it accelerates further with the initial velocity of 6 m/s. The rubber tyre has a mass of 1 kg. When another force of 5 N is applied on the tyre, its velocity increases and reaches up to ‘v’. Calculate the speed of a rubber tyre after 15 seconds.

**Solution:** **Given data:**

Force applied on the rubber tyre, F_{1} = 5 N

Initial velocity of the rubber tyre, u = 6 m/s

Mass of the rubber tyre, m = 1 kg

Force applied on the rubber tyre, F2 = 7 N

The net force applied on the rubber tyre,

F_{net} = (5 + 7)

F_{net} = 12 N

Final velocity of the rubber tyre, v =?

After time, t = 15 seconds**According to Newton’s second law equation,**

F_{net} = m (v – u) / t

12 = 1 (v – 6) / 15

(v – 6) = 12 × 15

(v – 6) = 180

v = 180 + 6

v = 186 m/s

Therefore, after 15 seconds the speed of the rubber tyre will be **186 m/s.**

**Numerical 5:**

One truck that has a mass of 975 kg is initially at rest. The truck attains the final velocity of 35 m/s in time 25 seconds. Calculate the total net force acting on the truck.

**Solution:** **Given data:**

Mass of the truck, m = 975 kg

Initial velocity of the truck, u = 0 m/s

Final velocity of the truck, v = 35 m/s

In time, t = 25 seconds

The net force acting on the truck, F_{net} =?**According to Newton’s second law equation,**

F_{net} = m (v – u) / t

F_{net} = 975 (35 – 0) / 25

F_{net} = 39 × 35

F_{net} = 1365 N

Therefore, a net force of **1365 N** is acting on the truck.

**........**

Don’t you think, is easy to remember all the three Newton’s second law equations?

(Let me know by leaving a **comment**)

If you want to **read more** about the Newton’s laws,

**You can check here:**

**Newton’s second law of motion**Newton’s second law example

Definition of newton’s second law

**Newton’s first law of motion**

Newton’s first law example

**Newton’s third law of motion**

Newton’s third law example

**Newton’s laws of motion**

How many newton’s laws are there

**Newton’s law of cooling**

Newton’s law of cooling formula

**Newton’s law of inertia**

Newton’s law of inertia examples

**Newton’s universal law of gravitation**