FORCE & PRESSURE 2
CHAPTER
NEWTON'S LAWS OF MOTION
CONTENTS
(A) Newton´s Ist law :
Force
A body can not change its state of motion by itself.
Newton's laws of motion If the object is at rest it will remain at rest and if
it is in uniform motion, it continues to be in
Thrust and pressure
motion unless some external force is applied on it.
Buoyancy This law is also known as law of inertia.
Density (B) Newton´s second law :
Newton's second law can be written as
v u
F ma m
t
FORCE Ex.1 Calculate the force required to produce an
acceleration of 5 m/s2 in a body of mass 2.4 kg.
The external agent which can changes :
Sol. We know that force = mass × acceleration
the speed and direction of motion or
= 2.4 kg × 5 m/s2 = 12.0N
the shape of a body
is called force. Ex.2 A force acts for 0.2 s on a body of mass 2.5
kg initially at rest. The force then ceases to
It is a vector quantity
act and the body moves through 4m in the
Types of forces :
next one second. Calculate the magnitude of
(A) Contact forces : force.
The forces that act on bodies when they are in Sol. When the force ceases to act, the body will
actual contact are known as contact forces. move with a constant velocity. Since it moves
Ex. Frictional force, normal reaction force, a distance of 4 m in 1 s, therefore, its uniform
tension in string, force exerted during velocity = 4m/s.
collision, force applied as a push or a pull etc. Now, initial velocity, u=0
Final velocity, v = 4 m/s
(B) Non-contact forces :
The forces that act on bodies without being Time interval t = 0.2 s
touched are called non-contact forces. vu 40
Acceleration, a = 20m/s2
Ex. Gravitational force, electrostatic force, t 0 .2
magnetic force etc. From the relation,
F = ma, we get
Force, F = 2.5 × 20 = 50 N
, Ex.3 A ball of mass 20 gm is initially moving with Therefore, the acceleration of the motorcycle,
a velocity of 100 m/s. On applying a constant 30
a 6 m/s2
force on the ball for 0.5s, it acquires a 5
velocity of 150 m/s. Calculate the following : The magnitude of the force applied by the
(i) Acceleration of the ball brakes is given by the equation,
(ii) Magnitude of the force applied F = mass × acceleration
Sol. Given , m = 20 gm = kg = 0.02 kg = 250 kg × (6)m/s2 = 1500 N
Initial velocity, u = 100 m/s
(C) Newton's third law of motion
Time interval, t = 0.5 s Newton's third law of motion states that " if a
Final velocity, v = 150 m/s body A exerts a force on the body B, the body
v u 150 100 B will also exert an equal and opposite force
(i) Acceleration, a = 100ms –2
t 0.5 on A."
(ii) Force, F = mass × acceleration Newton's third law is also stated as "to every
=0.02 × 100 = 2.0 N action there is an equal and opposite
reaction."
Ex.4 A cricket ball of mass 200 gm moving with a The force exerted by A on B is called action
speed of 40 m/s is brought to rest by a player while the force exerted by B on A is called
in 0.04s. Calculate the average force applied the reaction.
by the player. Action and reaction always act on different
200 bodies.
Sol. Mass, m = 200 gm = kg = 0.2 kg
1000 Forces always occur in pairs.
Initial velocity, u = 40 m/s eg. by hitting a table with palm we apply a
Final velocity, v = 0 force. The table also exerts a force on palm
Time, t = 0.04s on hitting it.
Average force =
Change in momentum 8.0
= – 200 N THRUST AND PRESSURE
Time 0.04
(The negative sign shows that the force is (A) Thrust :
applied in a direction opposite to the direction The force acting normally on surface is called
of motion of the ball). 'thrust'.
Ex. 5 A motorcycle is moving with a velocity of This is a vector quantity.
It is measured in newton (N).
108 km/hr and it takes 5 s to stop it after the
brakes are applied. Calculate the force (B) Pressures :
exerted by the brakes on the motorcycle if its The thrust on an unit area of a surface is
mass along with the rider is 250 kg. called 'pressure'.
Sol. Given that initial velocity of the motorcycle Thrust F
Pressure = or P
= 108 km/hr = 30 m/s Area A
Final velocity = 0 m/s Unit : The SI unit of pressure is newton per
Time taken to stop = 5s, the mass of the meter square or N/m2, other units of pressure
motorcycle with rider = 250 kg. are pascal and bar.
The change in the velocity of the motorcycle One Pascal : One pascal is defined as the
in 5s = 0 – 30 = –30 m/s pressure exerted on a surface area of 1m2 by a
thrust of 1 newton. i.e. 1 Pascal = 1 N/m2
CHAPTER
NEWTON'S LAWS OF MOTION
CONTENTS
(A) Newton´s Ist law :
Force
A body can not change its state of motion by itself.
Newton's laws of motion If the object is at rest it will remain at rest and if
it is in uniform motion, it continues to be in
Thrust and pressure
motion unless some external force is applied on it.
Buoyancy This law is also known as law of inertia.
Density (B) Newton´s second law :
Newton's second law can be written as
v u
F ma m
t
FORCE Ex.1 Calculate the force required to produce an
acceleration of 5 m/s2 in a body of mass 2.4 kg.
The external agent which can changes :
Sol. We know that force = mass × acceleration
the speed and direction of motion or
= 2.4 kg × 5 m/s2 = 12.0N
the shape of a body
is called force. Ex.2 A force acts for 0.2 s on a body of mass 2.5
kg initially at rest. The force then ceases to
It is a vector quantity
act and the body moves through 4m in the
Types of forces :
next one second. Calculate the magnitude of
(A) Contact forces : force.
The forces that act on bodies when they are in Sol. When the force ceases to act, the body will
actual contact are known as contact forces. move with a constant velocity. Since it moves
Ex. Frictional force, normal reaction force, a distance of 4 m in 1 s, therefore, its uniform
tension in string, force exerted during velocity = 4m/s.
collision, force applied as a push or a pull etc. Now, initial velocity, u=0
Final velocity, v = 4 m/s
(B) Non-contact forces :
The forces that act on bodies without being Time interval t = 0.2 s
touched are called non-contact forces. vu 40
Acceleration, a = 20m/s2
Ex. Gravitational force, electrostatic force, t 0 .2
magnetic force etc. From the relation,
F = ma, we get
Force, F = 2.5 × 20 = 50 N
, Ex.3 A ball of mass 20 gm is initially moving with Therefore, the acceleration of the motorcycle,
a velocity of 100 m/s. On applying a constant 30
a 6 m/s2
force on the ball for 0.5s, it acquires a 5
velocity of 150 m/s. Calculate the following : The magnitude of the force applied by the
(i) Acceleration of the ball brakes is given by the equation,
(ii) Magnitude of the force applied F = mass × acceleration
Sol. Given , m = 20 gm = kg = 0.02 kg = 250 kg × (6)m/s2 = 1500 N
Initial velocity, u = 100 m/s
(C) Newton's third law of motion
Time interval, t = 0.5 s Newton's third law of motion states that " if a
Final velocity, v = 150 m/s body A exerts a force on the body B, the body
v u 150 100 B will also exert an equal and opposite force
(i) Acceleration, a = 100ms –2
t 0.5 on A."
(ii) Force, F = mass × acceleration Newton's third law is also stated as "to every
=0.02 × 100 = 2.0 N action there is an equal and opposite
reaction."
Ex.4 A cricket ball of mass 200 gm moving with a The force exerted by A on B is called action
speed of 40 m/s is brought to rest by a player while the force exerted by B on A is called
in 0.04s. Calculate the average force applied the reaction.
by the player. Action and reaction always act on different
200 bodies.
Sol. Mass, m = 200 gm = kg = 0.2 kg
1000 Forces always occur in pairs.
Initial velocity, u = 40 m/s eg. by hitting a table with palm we apply a
Final velocity, v = 0 force. The table also exerts a force on palm
Time, t = 0.04s on hitting it.
Average force =
Change in momentum 8.0
= – 200 N THRUST AND PRESSURE
Time 0.04
(The negative sign shows that the force is (A) Thrust :
applied in a direction opposite to the direction The force acting normally on surface is called
of motion of the ball). 'thrust'.
Ex. 5 A motorcycle is moving with a velocity of This is a vector quantity.
It is measured in newton (N).
108 km/hr and it takes 5 s to stop it after the
brakes are applied. Calculate the force (B) Pressures :
exerted by the brakes on the motorcycle if its The thrust on an unit area of a surface is
mass along with the rider is 250 kg. called 'pressure'.
Sol. Given that initial velocity of the motorcycle Thrust F
Pressure = or P
= 108 km/hr = 30 m/s Area A
Final velocity = 0 m/s Unit : The SI unit of pressure is newton per
Time taken to stop = 5s, the mass of the meter square or N/m2, other units of pressure
motorcycle with rider = 250 kg. are pascal and bar.
The change in the velocity of the motorcycle One Pascal : One pascal is defined as the
in 5s = 0 – 30 = –30 m/s pressure exerted on a surface area of 1m2 by a
thrust of 1 newton. i.e. 1 Pascal = 1 N/m2
