Newton’s Second Law equation (15 Easy Solved problems)

A teacher explaining Newton's second law equation

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

  • Equation 1: [Fnet = ma]
  • Equation 2: [p= mv]
  • Equation 3: [Fnet = m(v-u) / t]

Before moving onto the numericals, 

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

(One by One)

Newton's second law equation, F=ma

Where,

Fnet = net force applied on the body, N

m = mass of the object, kg

a = acceleration of the object, m/s2

Newton's second law equation, p=mv

Where,

p = Momentum of the object, kg m/s

v = velocity of the object, m/s

Newton's second law equation, F=m(v-u) / t

Where,

m (v – u) / t = rate of change in momentum, kg m/s2

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 2nd 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/s2?

Solution:

Given data:

Mass of the ball, m = 4 kg

The net force required, Fnet =?

Acceleration of the ball, a = 9 m/s2

According to Newton’s 2nd law formula,

Fnet = ma

Fnet = 4 × 9

Fnet = 36 N

Therefore, a net force of 36 N is required to accelerate the ball at a rate of 9 m/s2.

Numerical 2:
If the object is accelerating forward at a rate of 10 m/s2, 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/s2

Force applied on the object, Fnet = 15 N

Mass of the object, m =?

According to Newton’s second law formula,

Fnet = ma

m = Fnet / 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/s2. 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/s2

Mass of the bicycle, m = 2.5 kg

The net force applied on the bicycle, Fnet =?

According to Newton’s second law formula,

Fnet = ma

Fnet = 2.5 × 3

Fnet = 7.5 N

Therefore, when the bicycle accelerates at a rate of 3 m/s2, a force of 7.5 N is applied to it.

Numerical 4:
A 20 kg box accelerates at a rate of ‘a’ m/s2. 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, Fnet = 600 N

Acceleration of the box, a =?

According to the formula of Newton’s second law,

Fnet = ma

a = Fnet / m

a = 600 / 20

a = 30 m/s2

Therefore, the box accelerates at a rate of 30 m/s2when a force of 600 N is applied on it.

Numerical 5:
When a 92 N force is applied on an object, it accelerates further at a rate of 4 m/s2. What is the mass of an object?

Solution:

Given data:

The net force applied on the object, Fnet = 92 N

Acceleration of the object, a = 4 m/s2

Mass of the object, m =?

According to Newton’s second law,

Fnet = ma

m = Fnet / 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, Fnet = 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,

Fnet = 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, Fnet = – 5 N

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

Time, t =?

As per the Newton’s second law,

Fnet = 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, Fnet = – 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 2nd law equation,

Fnet = 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, F1 = 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,

Fnet = (5 + 7)

Fnet = 12 N

Final velocity of the rubber tyre, v =?

After time, t = 15 seconds

According to Newton’s second law equation,

Fnet = 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, Fnet =?

According to Newton’s second law equation,

Fnet = m (v – u) / t

Fnet = 975 (35 – 0) / 25

Fnet  = 39 × 35

Fnet  = 1365 N

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

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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

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