In this article, we shall study to solve numerical problems to calculate the magnitude of force, momentum, and change in momentum.

Example – 01:

The speed of a tortoise and hare are 1 m/s and 3 m/s respectively. The mass of the hare is 5 kg while that of tortoise is 20 kg. Which of the two has greater momentum?

Solution:

The momentum of hare = Mass of hare x Speed of hare= 5 x 3 = 15 kg m/s

The momentum of tortoise = Mass of tortoise x Speed of tortoise= 20 x 1 = 20 kg m/s

Thus the momentum of tortoise is more than that of the hare.

Example – 02:

What is the magnitude of force exerted by a horse in pulling a cart of mass 600 kg and accelerating at the rate of 1.2 m/s²?

Given: mass of cart = m = 600 kg, acceleration = a = 1.2 m/s².

To Find: F =?

Solution:

By Newton’s second law of motion, magnitude of force

F = ma = 600  x 1.2

∴ F = 720 N

Ans: The required force is 720 N.

Example – 03:

An object of mass 10 kg is moving with the initial velocity of 10 m/s. A constant force acts on it for 4 s giving it a speed of 2 m/s in opposite direction. Find the acceleration and the force.

Given: mass of object = m = 10 kg, initial velocity = u = 10 m/s, Final velocity = v = – 2m/s (negative since it is in opposite direction), time for which force is acting = t = 4 s

To Find: acceleration = a =? Force acting = F =?

Solution:

a = (v – u)/t = (-2 – 10)/4 = -12/4 = -3 m/s²

The negative sign indicated retardation.

By newton’s second law of motion we have

F = ma = 10 x (-3) = -30 N

Negative sign indicated retarding force.

Ans: The acceleration = 3 m/s² and Force = -30 N

Example – 04:

A constant force of 2 N acts on a body for 5 seconds to change its velocity. Calculate the change in the momentum.

Given: Force acting = F = 2 N, time for which the force is acting = t = 5s.

To Find:  Change in momentum =?

Solution:

Change in momentum = F x t = 2  x 5  = 10 kg m/s.

Ans: the change in momentum is 10 kg m/s.

Example – 05:

An empty truck of mass 1000 kg is moving at a speed of 36 km/hr. It is loaded with 500 kg of material on its way and again moves at the same speed. Will the momentum of the truck remain the same after loading? if not, find the momentum of the truck after loading.

Given: Mass of a truck = 1000 kg, Mass of Load = 500 kg, Speed of the vehicle = v = 36 km/hr = 36 x 5/18 = 10 m/s.

To Find: Momentum of truck = p =?

Solution:

The momentum of a body depends on its mass. In this case, the truck is loaded on the way, hence its momentum should change.

Total Mass = m = 1000 + 500 = 1500 kg

New momentum = Total Mass x Velocity = 1500 X 10 = 15000 kg m/s

Ans: The momentum of the truck after loading is 15000 kg m/s.

Example – 06:

A railway wagon of mass 1000 kg is pulled with a force of 10000 N. What is the acceleration?

Given : Force applied = F = 10000 N, mass of wagon = m = 1000 kg.

To Find:  acceleration = a =?,

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴ 10000 = 1000 x a

∴ a = 10000/1000 = 10 m/s².

Ans: The acceleration is 10 m/s².

Example – 07:

A car of mass 1000 kg is moving at a certain speed when a constant braking force 1000 N acts on it for 5 s and speed of the car reduced to half the original speed. Find the further time required to stop the car, if the same constant force acts on it.

Given: mass of car = m = 1000 kg, Force acting = F = 1000 N,time taken =  t = 5 s, Final speed (v) = 1/2 Initial speed (u) = u/2

To find: t =? when v = 0

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴1000 = 1000 x a

∴ a = 1000/1000 = 1 m/s².

By the first equation of motion

∴ a = (v – u)/t  = (u/2 – u)/5 = (-u/2)/5 = (-u/10)

∴ a = -u/10 = – 1

∴ u = 10 m/s

We have v = u + at

∴ 0 = 10 + (-1)t

∴ -10 = – t

∴ t = 10 s

Ans: The further time required to stop the car is 10 s

Example – 08:

Find the magnitude of the force applied to a block of mass 5 kg at rest, if it moves 36 m, in first 3 seconds. Neglect the force of friction.

Given: mass of block = m = 5 kg, Initial velocity = u = 0, Distance traveled = s = 36 m, time taken = t = 3 s,

To find: F =?

Solution:

s = ut + 1/2 at²

∴   36 = (0)(3) + 1/2 a(3)²

∴   36 = 1/2 a(9)

∴   72 = a (9)

∴ a = 72/9 = 8 m/s².

By Newton’s second law of motion, the magnitude of force

F = ma = 5 x 8 = 40 N

Ans: The force applied = 40 N

Example – 09:

Two spheres of mass 10 g and 100 g each fall on the two pans of a table balance from a height of 40 cm and 10 cm respectively. If both are brought to rest in 0.1 seconds. Determine the force exerted by each sphere on the pans.

Solution:

For the first sphere: 

Given : m₁ = 10 g = 0.01 kg, h = – 40 cm  = – 0.4 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴  v² = (0)² + 2(-9.8)(-0.4)

∴  v² =  7.84

∴  v = 2.8 m/s

F₁ = m₁a

F₁ = m₁ (v – u)/t  = 0.01 x (2.8 – 0)/0.1 = 0.28 N

For the second sphere: 

Given : m₁ = 100 g = 0.1 kg, h = – 10 cm  = – 0.1 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴ v² = (0)² + 2(-9.8)(-0.1)

∴ v² =  1.96

∴ v = 1.4 m/s

F₂ = m₂a

F₂ = m₂ (v – u)/t  = 0.1 x (1.4 – 0)/0.1 = 1.4 N

Ans: The force exerted by the first sphere is 0.28 N and that by second sphere is 1.4 N

Example – 10:

Calculate the density of a cubical ice block of side 50 cm. If a force of 1125 N applied to it produces an acceleration of 10 m/s² in it. Neglect the force of friction. Assume the ice block remains in solid state without melting.

Given: Force applied = F = 1125 N, acceleration = a = 10 ms-1, Side of a block = 50 cm = 0.5 m

To Find:  Density = ρ = ?,

Solution:

By Newton’s second law of motion

F = m.a

∴ 1125 = m x 10

∴ m = 1125/10 = 112.5 kg

Ans: The density of ice block is 900 kg/m³.

Example – 11:

A ball of mass 50 g at rest is hit by a bat and the ball covers a distance of 400 m, in 2 seconds. If the ball was in contact with it for 0.1 s, find the magnitude of the force acting on it. Assuming no other force acts on a ball after it is hit by the bat.

Given: mass of ball = m = 50 g = 0.05 kg,  time of contact = 0.1 s, distance covered  s= 400 m, time taken to cover the distance = t = 2 s.

To find: Force acting =F =?

Solution:

No other force acts on a ball after it is hit by the bat. Thus it is in uniform motion after hit.

v = s/t = 400/2 = 200 m/s

Now Force, F = ma

F = m (v – u)/t  = 0.05 x (200 – 0)/0.1 = 100 N

Ans: The force acting = 100 N

Example – 12:

A force of 500 N acts on a body of mass 1000 kg and the body is brought to rest within a distance of 64 m. Find the initial velocity and time taken by the body to come to rest.

Given: mass of body = m = 1000 kg, Force acting = F = 500 N, Final velocity = v = 0 ms-1, distance traveled = s = 64m.

To find: initial velocity =u =? time taken = t = ?

Solution:

We have     F = ma

∴   500 = 1000 x a

∴ a = 500/1000 = 0.5 m/s²

As the body is brought to rest a = – 0.5 m/s²

v² = u² + 2as

∴ (0)² = u² + 2(-0.5)(64)

∴ (0)² = u² – (64)

∴ u² = 64

∴ u = 8 m/s

By first equation of motion

v = u + at

∴ 0 = 8 + (- 0.5) x t

∴ 8 = – 0.5 x t

∴ t = 8/0.5 = 16 s

Ans: Initial velocity = 8 m/s, time taken to come to rest = 16 s

Example – 13:

A car of mass 1000 kg is moving uniformly with 10 m/s. If the engine of the car develops an extra linear momentum of 1000 kg m/s. Calculate the new velocity with which the car runs.

Solution:

Given : mass of car = m = 100 kg, initial velocity = u = 10 m/s, Extra momentum = 1000 kg m/s,

To find: Final velocity = v = ?

Initial momentum = p₁ = mu = 1000 x 10 = 10000 kg m/s

Final momentum = p₂ = 10000 + 1000 = 11000 kg ms-1.

Now, Final momentum = p₂ = mv

∴ 11000 = 1000 x v

∴ v = 11000/1000 = 11m/s

Ans: New Velocity = 11 m/s

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Depending upon the interaction between the bodies and level of contact forces are classified a) Contact forces b) Non-contact forces. Muscular force is a contact force. Electrostatic force is non contact force.

Contact Forces:

A force which can be applied only when it is in contact with an object is called a contact force. All mechanical forces are contact forces. The following are the examples of contact force.

Muscular Force: The force resulting due to the action of muscles is known as muscular force. Muscular force can be applied only when it is in contact with the object. For ease, a rope or handle may be used. We use muscular force to different activities using hands like lifting, throwing, catching, etc. We use the muscular force of animals to simplify our work. We use instruments with the help of muscular force. Examples of the use of muscular force are cutting of vegetables or fruits using a knife, hammering of a nail, kneading of the dough, bullocks ploughing in the field, horse pulling a cart, etc.

Friction: The force of friction always acts on all the moving objects and its direction is always opposite to the direction of motion. It always acts a point of contact between the two bodies. i.e. It arises due to contact between the two surfaces. Examples: When a vehicle is moving on a road, there is a force of friction between the surface of the road and tyres at the point of contact. Other examples are a ball rolling on ground stops after some time due to friction between the ball and the ground. If we stop pedaling a bicycle. It stops after some distance. It is due to friction between the road surface and the tyres. It is difficult to slide heavier objects on the ground. After some time the tyres of vehicle wear and tear.

Types Of Contact Forces

There are 6 kinds of forces which act on objects when they come into contact with one another. Remember, a force is either a push or a pull.

• Normal Force: A book resting on a table has the force of gravity pulling it toward the Earth. But the book is not moving or accelerating, so there must be opposing forces acting on the book. This force is caused by the table and is known as the normal force.
• Applied Force: Applied force refers to a force that is applied to an object such as when a person moves a piece of furniture across the room or pushes a button on the remote control. A force is applied.
• Frictional Force: Frictional force is the force caused by the relative motion of two surfaces that come into contact with each other.
• Tension Force: Tension force is the force applied to a cable or wire that is anchored on opposite ends to opposing walls or other objects. This causes a force that pulls equally in both directions.
• Spring Force: The spring force is the force created by a compressed or stretched spring. Depending upon how the spring is attached, it can pull or push in order to create a force.
• Resisting Forces: The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude (and due to the fact that it is mathematically difficult to predict its value). It is most noticeable for objects that travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas.

Non-Contact Forces:

A force which can be applied without any contact with two bodies is called a non-contact force. The following are examples of non-contact force.

• Magnetic Force: The force exerted by a magnet is called a magnetic force e.g. Like poles of a magnet repel each other, while unlike poles attract each other. These forces exist without contact between the two magnets. e.g. i) Magnet attracts Cobalt, Nickel, Iron and steel towards itself. ii) When two magnets are brought near then South-South, North – North poles repel each other. While North and South pole attracts each other.
• Electrostatic force: The force exerted by a charged body is called an electrostatic force e.g. Like charges repel each other, while unlike charges attract each other. These forces exist without contact between the two bodies. e.g. i) When a comb is run through dry hair, then the comb can attract small pieces of paper. ii) If we bring a charged comb near our hair, they rise towards the comb. iii) A charge is developed on synthetic fibre due to rubbing. iv) We can hear the cracking sound of sparks when taking off or putting on woolen or synthetic clothes.
• Gravitational force: The force exerted by the earth on a body is called gravitational force. Actually this force exists between any two bodies in the universe. This force is always of attraction. e.g. When a body is dropped from a height it moves in a downward direction towards the Earth with increasing speed (with constant acceleration). This constant acceleration by which all bodies fall down is called acceleration due to gravity. Its value is 9.8 m/s² (approx 10  m/s² )on the surface of the earth. e.g. i) A fruit from tree falls down; ii) Water falls down on the ground from a tap. iii) We feel the weight of a bucket full of water holding in our hand.

Force field:

Region or space in which non-contact force such as magnetic force, gravitational force, electrostatic force acts is called force field. The region surrounding a magnet, where magnetic substance experiences force is called the magnetic field. The region surrounding an electric charge, where electric charge experiences force is called the electric field. Thus a field is a sphere of influence of non-contact force.

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