Class 9 Science

# Newton's Second Law of Motion

## Numerical Problems

### Part 2

Type - IV

Question 1: An object of 50 kg gets the speed of 10m/s in 5 second from zero velocity. Calculate the required force applied by engine of the car.

Solution: According to the question:-

Initial velocity (u) = 0, final velocity (v) = 10m/s, time (t) = 5 second, Mass (m) = 50 kg,
Therefore, force (F)=?
We know that, Force (F) =m(v-u)/t
:. F=50\ kg\ (10\ m//s - 0)/(5\ s)
=> F = 50\ kg xx 2\ ms^(-2)
=> 100\ kg\ ms^(-2) => 100N
Thus required force = 100 N

Question 2: A car having mass of 1500 kg achieve the velocity of 5 m/s in 10 second. Calculate the required force to attain required speed by car.

Solution: According to the question:

Initial velocity (u) = 0, final velocity (v) = 5m/s, time (t) = 10 second, Mass (m) = 1500 kg,
Therefore, force (F)=?
We know that, Force (F) =m(v-u)/t
:. F=1500\ kg\ (5\ m//s - 0)/(10\ s)
=> F = 1500\ kg xx 1/2\ ms^(-2)
=> 750\ kg\ ms^(-2) => 750N
Thus required force = 750 N

Question 3: A bus starts from the stop and take 50 second to get the speed of 10m/s. If the mass of the bus along with passengers is 10000 kg, calculate the force applied by the engine of bus to push the bus at the speed of 10m/s.

Solution: According to the question:

Initial velocity (u) = 0, final velocity (v) = 10m/s, time (t) = 50 second, Mass (m) = 10000 kg,
Therefore, force (F)=?
We know that, Force (F) =m(v-u)/t
:. F=10000\ kg\ (10\ m//s - 0)/(50\ s)
=> F = 10000\ kg xx 1/5\ ms^(-2)
=> 2000\ kg\ ms^(-2) => 2000N
Thus required force = 2000 N

Question 4: An object gets 50 second to increase the speed from 10m/s to 50m/s. If the mass of the object is 1000 kg, what force will be required to do so?

Solution: According to the question:

Initial velocity (u) = 10m/s, final velocity (v) = 50m/s, time (t) = 50 second, Mass (m) = 1000 kg,
Therefore, force (F)=?
We know that, Force (F) =m(v-u)/t
:. F=1000\ kg\ (50\ m//s - 10\ m//s)/(50\ s)
=> F = 1000\ kg xx 40/50\ ms^(-2)
=> F= 20\ kgxx40 ms^(-2)
=>F=800\ kg\ ms^(-2) = 800N
Thus required force = 800 N

Question 5: What force will be required to speed up a car having mass of 1200kg, from 5 m/s to 15m/s in 10 second?

Solution: According to the question:-

Initial velocity (u) = 5m/s, final velocity (v) = 15m/s, time (t) = 10 second, Mass (m) = 1200 kg,
Therefore, force (F)=?
We know that, Force (F) =m(v-u)/t
=>F=1200\ kg\ (15\ m//s - 5\ m//s)/(10s)
=>F=1200\ kg xx (10\ m//s)/(10s)
=>F=1200\ kg xx 1\ ms^(-2)
=>F=1200\ kg\ ms^(-2) = 1200N
Thus required force = 1200 N

Question 6: In how much time an object having mass of 100kg will speed up from 5m/s to 25m/s, if 500N force will be applied over it?

Solution: According to the question:

Initial velocity (u) = 5m/s, final velocity (v) = 25m/s, Mass (m) = 100 kg, Force (F) = 500N
Therefore, time (t) = ?
We know that, Force (F) =m(v-u)/t
:. 500N = 100\ kg\ (25m//s-5m//s)/t
=>500N=100\ kg xx (20m//s)/t
=>500N xx t=2000\ kg\ m//s
=>t= (2000\ kg\ m//s)/(500\ kg\ m//s^2) = 4s
Thus required time = 4 second

Question 7: If a force of 1000 N is applied over a vehicle of 500 kg, then in how much time the speed of the vehicle will increase from 2 m/s to 10 m/s?

Solution: According to the question:

Initial velocity (u) = 2m/s, final velocity (v) = 10m/s, Mass (m) = 500 kg, Force (F) = 1000N
Therefore, time (t) =?
We know that, Force (F) =m(v-u)/t
:. 1000N = 500\ kg\ (10m//s-2m//s)/t
=>1000N=500\ kg xx (8m//s)/t
=>1000N xx t=4000\ kg\ m//s
=>t= (4000\ kg\ m//s)/(1000\ kg\ m//s^2) = 4s
Thus required time = 4 second

Question 8: A vehicle having mass equal to 1000 kg is running with a speed of 5m/s. After applying the force of 1000N for 10 second what will be the speed of vehicle?

Solution: According to the question:

Mass of (m) = 1000 kg, Force, (F) = 1000 N, time (t) = 10s, Initial velocity (u) = 5m/s
Therefore, Final velocity (v) =?
We know that, Force (F) =m(v-u)/t
:. 1000N = 1000\ kg\ (v-5m//s)/(10s)
⇒ 1000 kg m/s2 × 10s = 1000 kg (v – 5m/s)
⇒ 10000 kg m/s = 1000 kg × v – 5000 kg m/s
⇒ 10000 kg m/s + 5000kg m/s = 1000kg × v
⇒ 15000 kgm/s = 1000 kg × v
=>v=(15000\ kg\ m//s)/(1000\ kg)=15\ m//s
Thus, the velocity of the vehicle will be 15m/s.

Question 9: An object gets the velocity of 10 m/s after applying a force of 500N for 10 second. If the mass of the object is equal to 1000 kg, what was its velocity before applying the force?

Solution: According to the question:

Mass (m) = 1000 kg, Force (F) = 500N, time (t) = 10m/s, Final velocity (v) = 10m/s
Therefore, Initial velocity (u) =?

We know that, Force (F) =m(v-u)/t
=>500N = 1000\ kg\ (10m//s-u)/(10s)
=>500\ kg\ ms^(-2) = (1000\ kgxx10m//s-1000\ kg xx u)/(10s)
⇒ 500 kg ms–2 × 10s = 10000 kg ms–1–1000 kg × u
⇒ 5000 kg m/s = 10000 kg m/s – 1000 kg × u
⇒ 5000 kg m/s – 10000 kg m/s = –1000 kg × u
⇒ –5000kg m/s = –1000 kg × u
=>u=(-5000\ kg\ m//s)/(-1000\ kg)=5m//s
Thus speed of object was 5m/s

Type V:

Question 1: The acceleration of two objects are 5m/s2 and 20m/s2. If mass of both the object would be combined and a force of 50N would be applied on them, what will be their acceleration?

Solution: In the order to calculate the acceleration of both the objects after combining their mass, first of all their mass will be calculated.

Ist object:
Given, Acceleration (a) = 5m/s2
Let the mass of one body = m1
And a force of 50N will be applied over it.
We know that Force (F) = Mass (m) x Acceleration (a)
⇒ 50N = m1 × 5ms-2
=>m_1=(50N)/(5ms^(-2))=10\ kg

2nd Object:
Given, Acceleration (a) = 20m/s2
Let the mass of one body = m2
And a force of 50N will be applied over it.
We know that Force (F) = Mass (m) x Acceleration (a)
⇒ 50N = m2 × 5 ms-2
=>m_2=(50N)/(20ms^(-2))=2.5\ kg
Now their total mass = m1 + m2 = 10 kg + 2.5 kg = 12.5 kg

In this condition:
Mass (m) = 12.5 kg, Force (F) = 50N, therefore, Acceleration (a) =?
We know that, F = m x a
=>a = (50N)/(12.5\ kg) = 4\ ms^(-2)
Therefore, 50N = 12.5kg × a
Thus, Acceleration = 4 ms–2