Subject: Science

Simply, pull or push is called force. A book lying on the table continues to lie on the table unless an external force is applied on it. So, generally force is an external agency which changes or tends to change the state of a body from rest to motion or vice versa. The SI unit of force is newton (N). In CGS system, the unit of the force is dyne. 1 newton= 10^{5} dynes.

Consider a situation in which football is rolling in the ground. Ball stops rolling after some time. If a ground is perfectly smooth then ball will continue rolling forever in same direction with constant speed. It is a roughness of the surface that provides opposing force, which is called friction, to slow down the speed of the ball and finally stop the ball. Friction can be defined as the opposing force which acts on the opposite direction of motion.

Force acting on a body can do following things:

- It can change the speed of the body.
- It can change the direction of motion of the body.
- It can change the shape of the body.
- It can change the position of the body.

When a number of forces acting on a body do not change its states of rest or uniform motion in a straight line, the force are said to be balanced forces. For example, in the game tug of war, a game where two teams pull a rope in opposite direction. In this game rope remains in steady and does not move in any direction when two teams exert equal and opposite force in a rope.

When a number of forces acting on a body change its state of rest or uniform motion in a straight line, the forces is said to be unbalanced forces.

For example, if we push a toy car towards east, it moves towards east. This unbalanced force produces motion.

- Force is an external agency which changes or tends to change the state of a body from rest to motion or vice versa.
- When a number of forces acting on a body do not change its states of rest or uniform motion in a straight line, the force are said to be balanced forces.
- When a number of forces acting on a body change its state of rest or uniform motion in a straight line, the forces is said to be unbalanced forces.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Force is defined as the push or pull on a body which changes or tends to change the state of rest or uniform motion in a straight line. Its SI unit is Newton and CGS unit is dyne.

Force can affect the body in the following ways:

- It can change the shape of the body.
- It can change the direction of motion of the body.
- It can change the speed of the body.

When a number of forces acting on a body changes its state of rest or uniform motion in a straight line, then the forces are said to be unbalanced forces.

It is defined as the force which when applied on a body of 1kg mass produces an acceleration of 1m/s^{2}.

Solution:

Given,

Retarding force (f) = -10N

Mass (m) = 20kg

Initial velocity (u) = 10m/s

Time (t) = ?

Final velocity (v) = ?

By formula,

F = ma

or, a = F/m

or, a = -10/20 = -0.5 m/s^{2}

And,

a =

or, t =

or, t = or, t = 20s Hence, the body takes 20 s to stop.

Given,

Retarding force (f) = -10N

Mass (m) = 20kg

Initial velocity (u) = 10m/s

Time (t) = ?

Final velocity (v) = ?

By formula,

F = ma

or, a = F/m

or, a = -10/20 = -0.5 m/s

And,

a =

or, t =

or, t = or, t = 20s Hence, the body takes 20 s to stop.

When a number of forces acting on a body do not change its states of rest or uniform motion in a straight line, the force are said to be balanced forces.

Action and reaction are equal in magnitude and opposite in direction.Then why they don't cancel each other?

Though action and reaction are equal in magnitude and opposite in direction, they act on different bodies. Thus the forces acting on different bodies don't cancel each other. Hence, as they act on different bodies, action and reaction don't cancel each other.

A boy runs from his home to school at a speed at a speed of 3 \(ms^{-1}\) on a straight road. He walks back to his home at a speed of 5 \(ms^{-1}\). FInd his (i) average speed and (ii) average velocity.

A) Let d the distance between his home and school in a straight line.The time taken by the boy to run from home to school (t1) = \(\frac{d}{3}\)

(speed= \(\frac{distance}{time}\))

The time taken by him to walk from school to home (t2) = \(\frac{d}{5}\)

Total time taken (t) = t1+t2

= \(\frac{d}{3}\) + \(\frac{d}{5}\) = \(\frac{8}{15}\) d

Total Distance Travelled = d+d =2d

Average speed = \(\frac{Total Distance}{Total Time}\)

= \(\frac{2d}{8d}\) x 15

= 3.75 m/s

So,Average speed = 3.75 m/s

Again,

Since the boy comes back to the starting point (i.e.,home) his total displacement is zero.

Average velocity = \(\frac{Total Displacement}{Total Time}\)

= \(\frac{s}{t}\) = \(\frac{0}{8}\) = 0

Average velocity = 0m/s

The Distance between Janakpur and Kathmandu is 400km.A bus takes 8 hours to cover this distance.Find the average speed.

Total distance covered (s) = 400 km = 3x = \(10^{5}\) m [1km=1000m]

Total time taken (t) = 8 hours = 8 x 3600 s [ 1hr=3600sec]

We know that,

Average speed (V) = = \(\frac{Total\;distance\;covered (S) }{Total\;Time\;Taken(T) }\)

= \(\frac{4 X 10{^5}} {8 X 3600}\)

= 13.89 m/s

Therefore,the average speed of the bus is 13.89 m/s.

If a car travelling at a velocity of 6m/s along a straight line speed up uniformly to the velocity 8m/s in 2 s, find the acceleration.

Here, initial velocity (u) = 6m/s

Final Velocity = 8m/s

Time Taken(t) = 2s

Acceleration(a) = ?

We know that,

a= \(\frac{v-u}{t}\)

= \(\frac{8-6}{2}\)

= \(\frac{2}{2}\) = 1 m/\(s^{2}\)

Therefore, the acceleration of a car is 1 m/\(s^{2}\)

A bus starts from rest. If the acceeration of the bus is \(0.5 m/s^{2}\), what will be its velocity at the end of 2 minutes and what distance will it cover during that time ?

Here, Initial velocity (u) = 0

Acceleration (a) = \(0.5 m/s^{2}\)

Time Taken (t) = 2 minutes = 120 s [1 min=60 sec]

Final Velocity (V) = ?

Distance Covered (S)= ?

We have,

V=u+at

or, v = 0+0.5 x 120

v= 60m/s

Again,

s= \(\frac{u+v}{2}\)

= \(\frac{0+60}{2}\) x 120

= 3600m

Therefore, the final veocity of the bus is 60m/s and it will cover 3600m in 2 minutes

A ball is thrown upwards with a velocity of 40m/s.Calculate (i) the maximum height travelled by it and (ii) the time taken for it to return the initial position.

[**Take g = 10m/s\(^2\)]**

Here, Initial Velocity (u) = 40m/s

For upwoard direction, acceleration (a) = -10m/\(s^{2}\)

The velocity of the ball goes on decreasing as the ball attains height. At the highest point, final velocity (v)=0

We have,

v=u+at

or,0=40+(-10)t

or 10t= 40

or t = \(\frac{40}{10}\)

t= 4s

Again, we have

s= ut+\(\frac{1}{2}\) \(at^{2}\)

or,s= 40 x 2 + \(\frac{1}{2}\) (-10) (\(2^{2}\))

s= 80 + \(\frac{1}{2}\) x (-40)

or, s = 80-20 = 60 m

Since 2s is the time taken by the ball to reach the maximum height, so time taken to return to the initial position is 2t= 2x2 = 4s

Therefore, the maximum height attained is 60 m and the time taken for it to return to the initial position is 4s.

A car is travelling at a speed of 60 km/hr. On seeing a baby 20m ahead on the road , the driver of the car applies brakes and the car stops at a distance of 15m. What is its retardation and how long time does it take to come at rest ?

Although the driver sees a baby 60 m ahead on the road , he stops the car at a distance of 15 m, hence,

Distance covered (s) = 15m

Initial Velocity (v) = 0

90km/hr = \(\frac {60 \times 1000m }{60 \times 60s }\)

= 16.66 m/s

Retardation = ?

Time taken (t) = ?

We know ,

\(v^{2}\) =\(u^{2}\) + 2as

or, \(0^{2}\) = ( \(16.66^{2}\) ) + 2a\(\times\) 15

or, -625 = 30a

or, a = \(\frac{277.5}{30}\)

\(\therefore\) a= -9.25 m/\(s^{2}\)

Again , we have

v= u + at

or, 0 = 16.66 + (-20.83) t

or, -16.66 = -9.25t

or, t = \(\frac{16.66}{9.25}\)

\(\therefore\) t= 1.80 s

Therefore, the acceleration of the car is -9.25 m/\(s^{2}\). i.e. its retardation 9.25 m/\(s^{2}\) and it takes 1.8 to come at rest.

A cricket ball if mass 100g moving with a velocity of 15 m / s is brought to rest by a player in 0.05s . What is the average force applied by the player ?

Here Mass (m) = 100g = \(\frac{100}{1000}\) kg = 0.1

Initial Velocity (u) = 15m/s

Final Velocity (v) = 0

Time Taken (t) = 0.05s

We have,

v= u + at

or, 0 = 15 + a \(\times\) 0.05

or, a = \(\frac{15}{0.05}\)

\(\therefore\) a = -300 m/\(s^{2}\)

Again, we have

F= ma

or,F= 0.1 \(\times\) (-300)

\(\therefore\) = -30N

A force acting on a body of mass 100 g displaces it through 200 cm in 5 seconds. Find the magnitude of the force if the initial velocity of the body is zero.

Here,

Mass (m) = 100g = \(\frac{100}{1000}\) kg

= 0.1kg

Distance Covered (s)= 200cm = 2m

Time Taken (t) = 5s

Initial Velocity (u) = 0

We know,

\(\therefore\) s= ut + \(\frac{1}{2}\) \(at^{2}\)

or, 2= 0 \(\times\) t + \(\frac{1}{2}\) a \( (5)^{2}\)

a= 2 \(\frac{4}{25}\) m/ \(s^{2}\)

Again, we have

\(\therefore\) F = ma

or, F= (0.1) \(\times\) \(\frac{4}{25}\) = 0.016 N

Thus, the magnitude of the force is 0.016 N

A bus starts from rest. If the acceleration of the bus is 0.5m/s \(^2\), what will be its velocity at the end of 2 minutes and what distance will it cover during that time ?

[**Take g = 10m/s\(^2\)]**

Solution:

Here, Initial Velocity (U) = 0

Acceleration(a) = 0.5m/s \(^2\)

Time taken (t) = 2 minutes = 120 s

Final velocity (V) = ?

Distance Covered(s) = ?

We have,

v = u + at

or, v = 0 + 0.5 \(\times\) 120

v = 60m/s

Again,

s = \(\frac{u\;+v}{2}\) \(\times\) t

= \(\frac{o\;+60}{2}\) \(\times\) 120 = 3600 m

Therefore, the final velocity of the bus is 60m/s and it will cover 3600 m in 2 minutes.

2. A ball is thrown upwards with a velocity of 20m/s. Calculate (i).the maximum height travelled by it and (ii).the time taken for it to return the initial position.

Here, Initial velocity (u) = 20m/s

Acceleration (a) = -10m/s \(^2\)

Final Velocity (v) = 0

We have,

v = u + at

or, 0 = 20 + (-10) t

or, 10t = \(\frac{20}{10}\)

\(\therefore\) t = 2s

Again, we have

s = ut + \(\frac{1}{2}\) \(at^{2}\)

or, s = 20 \(\times\) 2 + \(\frac{1}{2}\) (-10) (2^{2)}or, s = 40 + \(\frac{1}{2}\) \(\times\) (-40)

or, s = 40 - 20 = 20m

Since, 2s is the time taken by the ball to reach the maximum height, so time taken to return to the initial position is 2t = 2 \(\times\) 2s = 4s.

Therefore, the maximum height attained is 20 m and the time taken is 20m and the time taken for it to return to the initial position is 4s.

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