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]]>Solution:

Data;

Force exerted = F = 40N

length of strip of grass = 20m

Angle = q = 20^{0}to find:

work = W =?

Calculation:

we know that

W =F.d

= Fd cosq

= 40(20) cos20^{0}

W = 800 x 0.93

**W = 7.51 x 10 ^{2}J**

Solution:

Data;

mass of the car = m = 800 kg

initial velocity = v_{i }= 54kmh^{-1} = v_{i} = 54 x 1000/ 60 x60

initial velocity = v_{i }= 15ms^{-1}

final velocity = v_{f }= 0

distance covered by car = d = 60

To find:

average retarding force = F = ?

What happened to the original kinetic energy =?

Calculation;

according to the work energy principle

Fd = 1/2mv_{f}^{2 }– 1/2mv_{i}^{2} Fx60 = 1/2x800x(0)^{2}– ½(800)(15)

F = 0- 400 x 225/60

**F = -1500N**

Solution:

Data;

mass of automobile = m = 1000kg

f = 480N

length = l = 100m

to find :

final velocity = v = ?

calculation;

we know that

K.E =1/2 mv^{2}

P.E = mgh

according to the law of conversion of energy

loss of P.E = work done against friction = gain in K.E

so we write

mgh = f x l = 1/2mv^{2} substituting the value we have

1000 x 9.8 x 10 – 480 x 100 = 1/2 x 1000 x v^{2}

98000 – 48000 = 500v^{2}100 =v^{2}**v = 10 ms ^{-1}**

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- 1st Year Physics Chapter 1 Solved Numericals Notes PDF
- FSC 1st Year Physics Chapter 2 Solved Numericals Notes PDF
- FSC 1st Year Physics Chapter 3 Numericals Notes

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]]>When force acts on a body and causes a displacement in the body work said to be done.

OR

The product of the magnitudes of the displacement and the component of force in the direction of displacement called work

W = F.d

W = Fd cos

The space around the earth in which its gravitational force acts on a body called gravitational field.

Work Done In Gravitational Field

when an object moved in the gravitational field, the work done by gravitational force. If the displacement is in the direction of gravitational force, the workk is positive. However, if the direction of displacement is opposite to that of the direction of the gravitational force, then the work is negative. If displacement is perpendicular to force, work done is zero.

Consider an object of mass m being displaced with constant velocity from point A to B along different paths in the presence of the gravitational force as shown in figure. In this case the gravitational force is equal to the weight mg of the object.

let the object moved through path 1. Work done by the gravitational force along the path ADB split into two parts.

the work done along the path AD is zero because the weight mg is perpendicular to this path

the work done along the path DB is mgh because the direction of the weight mg is opposite to that of the direction of the displacement h of this path

W_{DB }= mgh

Hence total work done in moving the object from A to B along path 1 is

W_{ADB }= W_{AD }+ W_{DB }= 0 + (-mgh)

W_{ADB }= -mgh

FSC 1st Year Physics Chapter 1 Notes pdf download

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