# Newton’s Second Law of Motion [Equation/Formula + Problems]

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)

Where,

Fnet = net force applied on the body, N

m = mass of the object, kg

a = acceleration of the object, m/s2

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/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)

Contents

## 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?

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