Class 9 Science Chapter 10 Important Extra Questions

W = Fs
Work is done when
(a) a force acts on an object
(b) object is displaced.

(a) When force acts in the direction of motion of body work done is positive. When a lawn roller is pulled forward, work done is positive.
(b) When force is perpendicular to the direction of motion, work done is zero. A porter carrying load does no work.

The energy required for the various life process comes from the food that we eat.

Zero.

Moment of a force produces rotatory motion in a body whereas when a forced does work on a body it produces translatory motion in it.

Work done by the coolie is zero, as W = Fs cos90⁰ = 0.

Zero because there is no displacement.

Law of conservation of energy.

The law of conservation of energy states that energy can neither be created, nor destroyed, it can only be transformed from one form to another.

Kinetic energy and potential energy.

By a factor of 4 (KE ∝ v²).

Work done against friction is a negative work.

The lighter body because v = √(2KE/m)

Elastic potential energy.

No, because both m and v² are always positive.

Chemical energy gets converted into electrical energy

Electric motor.

Heat energy is converted into electrical energy.

Potential energy (= mgh) does not change as it does not depend on velocity.

1800.

When the student is picking up books from the floor in the beginning.

(a) When a body falls, its potential energy gradually gets converted into kinetic energy. On reaching the ground, the whole of the potential energy of the body gets converted into kinetic energy.
(b) The commercial unit of energy is kilowatt hour [kWh]
1kWh is the energy used in one hour at the rate of 1000Js-1.

We know that
1kWh = 1kW × 1h
= 10000 W × 36000s
= 1000 Js-1 × 36000s
= 36000000J
So, 1kWh = 3.6 × 10⁶ J

Suppose a body of mass m is lifted vertically upwards through a distance h. In this case, the force required to lift the body will be equal to weight of the body, mg (where m is mass and g is acceleration due to gravity), Now,
Work done in lifting a body = Weight of body × Vertical distance
W = mg × h
= mgh
Where W = Work done, and h = Height through which the body is lifted.

Power is the rate of doing work. Kilowatt is the unit of power and kilowatt hour is the unit of energy.

Suppose a body of mass m is moving with velocity v. It is brought to rest by applying a retarding force F. Suppose it traverses a distance s before coming to rest.
Kinetic energy of body, KE = Work done by retarding force to stop it.
i.e. Kinetic energy = F.s …. (i)
But Retarding force, F = ma ……(ii)
Initial velocity = v, final velocity = 0
From equation v² = u² + 2as, we have
O = v² – 2as (because here a is retardation)
Distance s = v²/2a …..(iii)
Substituting value F and s from (ii) and (iii) in (i) we get
Kinetic energy, KE = ma × v²/2a = 1/2mv²

(a) (i) A cricket/hockey ball which has been hit hard and is travelling fast.
(ii) A shot put thrown by an athlete.
(b) Let the body of mass m at height h above the ground starting from rest, be falling freely.
Total energy of the body at height h
= mgh (PE) + 0 (KE) = mgh
After the body has fallen freely through a distance x (say),
KE = mgh, PE = mg (h –x)
Total energy = KE + PE
= mgh + mg (h – x) = mgh
When it reaches the ground KE =1/2 m.2gh = mgh
PE = 0 Total Energy = KE + PE = mgh
Thus the total mechanical energy, which is the sum of KE and PE is always equal to mgh.

(i) Mass, m = 2.0 kg, height h = 4.8 m
Change in gravitational potential energy = mgh
= 2 × 10 × 4.8 = 96 J
(b) Potential energy → Kinetic energy → Heat and/or sound energy
(ii) Weight, mg = 520 N, height h = 3.0 m, time, t = 5.0s
Useful power = (Work done )/(Time taken)
= (mg)h/t = (520×3 )/5
= 312 W

When an object is raised through a certain height above the ground, its energy increases. This is because the work is done on it, against gravity. The energy present in such an object is called gravitational potential energy. Thus the gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity. Consider a body of mass m lying at point P on the earth surface where its potential energy is taken as zero. As weight, mg acts vertically downwards so to lift the body to another position Q at a height h, we have to apply a minimum force which is equal to mg in the upward direction. Thus work is done on the body against the gravity.
We know that,
Work done, W = Fs ….(i)
As F = mg and s = h
Putting these value in equation (i), we get
W = mg × h = mgh
This work done on the body is equal to the gain in energy of the body. This is the potential energy of the body.
Potential energy PE = mgh

The mechanical energy (kinetic energy + potential energy) of a freely falling object remains constant. It may be shown by calculation as follows:
Suppose a body of mass m falls from point A, which is at height H from the surface of earth. Initially at point A, kinetic energy is zero and the body has only potential energy.
Total energy of body at point A
= Kinetic energy + Potential energy
= 0 + mgH = mgH …..(i)
Suppose during fall the body is at position B. The body has fallen at a distance x from its initial position. If velocity of body at B is v, then from formula v² = u² + 2as we have
v² = 0 + 2gx = 2gx
Kinetic energy of body at point B = 1/2 mv²
= 1/2 m × 2gx = mgx
Potential energy of point B = mg (H – x)
Total energy of body at point B = Kinetic energy + Potential energy
= mgx + mg(H – x) = mgH ……(ii)
Now suppose the body is at point C, just above the surface of earth(i.e. just about to strike the earth) Its potential energy is zero.
The height by which the body falls = H
If v is velocity of body at C, then from formula
v² = u² + 2as
We have u = 0, a = g, s = H
So, v² = 0 + 2gH = 2gH
Kinetic energy of body at position C = 1/2 mv²
= 1/2 m × 2gH = mgH
Total energy of body at C
= Kinetic energy + Potential energy
= mgH + 0 = mgH …..(iii)
Thus we can see that the sun of kinetic energy and potential energy of freely body at each point remains constant.
Thus under force of gravity, the total mechanical energy of body remains constant.

Initial velocity = v, then final velocity v’ = 3v
Initial kinetic energy = 1/2 mv²
Final kinetic energy (KE) = 1/2 mv² = 1/2 m(3v)² = 9( 1/2 mv² )
(KE)initial : (KE)final = 1: 9.

v² – u² = 2as
This gives, s = (v² – u²)/2as
F = ma
We can write work done (W) by this force F
W = Fs
W = ma ((v² – u² )/2as)
= 1/2 m (v² – u²) = 1/2 mv² – 1/2 mu²
= Final KE – Initial KE

Linear momentum of first object P₁ = m₁v₁ and of second object p₂ = m₂v₂
But p₁ = p₂
or, m₁v₁ = m₂v₂
If m₁ < m₂ then v₁ > v₂
(K.E.)₁ = 1/2 (m₁v₁) v₁ = 1/2 p₁v₁
and (K.E)₂ = 1/2 (m₂v₂)v₂ = 1/2 p₂v₂
So, ((KE)₁)/((K.E)₂ ) = (1/2 p₁v₁)/(1/2 p₂v₂ ) = v₁/v₂
But v₁ > v₂
Therefore, (K.E)₁ > (K.E)₂

(a) F = 250kg × 10ms-² (g = 10ms-²)
= 2500N
s = 1 m
W = F.s
= 2500 N × 1m
= 2500 Nm = 2500J
(b) Zero as the box does not move at all while holding it.
(c) In order to hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force, muscular effort is involved. So, they feel tired.

(i) Kinetic energy = 1/2 mv²
= 1/2 × 900 × (30)² = 1/2 × 900 × 900
= 4,05,000J
(ii) Energy used = Work done against resistive force
= Force × Distance
= 2,000 × 30 = 60,000J = 60kJ
(iii) Minimum power = (Energy used )/(Time taken )
= 60,000J/1s = 60,000W = 60Kw
(iv) Chemical energy.

Here, force F = 20N, displacement, s = 1m
Angle between force and displacement ∅ = 600.
Work done, W = Fs cos ∅ = 20 × 1 × cos 600 = 20 × 1×1/2 = 10J.

Work in lifting the mass is done against gravity.
Therefore, the work done is W = mgh
We know, m = 200kg, g = 10m/s² and h = 6m
W = mgh
= 200 × 10 × 6
= 12000J

Here mass of the body m = 100g = 0.1 kg
Kinetic energy, K.E. = 20J = 20Nm = 20 kg
Kinetic energy = 1/2 mv²
v = √{(2 × kinetic energy)/(m )}
= √{(2 × 20kg m²/s² )/(0.1kg)}
= 20m/s
So the velocity of the body is 20m/s.

Given mass, m = 500kg
The given unit of speed is km/h. It is converted into m/s
1 km/h = (1 × 1000meter )/(3600 second) = (5)/(18)m/s
Initial speed u, = 36km/h = 36 × (5)/(18)m/s = 10m/s
Final Speed, v = 72 km/h = 72 × (5)/(18)m/s = 20m/s
Gain in KE = Final KE – Initial KE
= 1/2 mv² – 1/2 mv² = 1/2 m × (v² – u²)
= 1/2 × 500 × [(20)² – (10)²]
= 1/2 × 500[400 – 100]
= 1/2 × 500 × 300
= 75000joule
= 7.5 × 10⁴ J

Height of each step = 20cm = 0.20cm
Height of the 30 steps, H = 30 × 0.20
Time t = 15 seconds
Force exerted by man against gravity
F = Weight of man = mg
Work done by man = Force × Height = mg × H
Power P = (Work)/(Time) = (mgH )/(t)
= (60 ×10 ×6.0)/(15) = 240Joule.

Given force, F = 750N, displacement = height h = 16m and time t = 5
Work done by machine,
W = Force × Displacement = F × h
= 750N × 16 Joules
Power of machine, P = W/t = (750N×16J )/(5s ) = 2400 watt.

Given power, P = 40kW = 40 × 10³ = 40,000W
Force applied, F = Load = 20,000N
If v is the speed of load, then we have
P = Fv
Speed v = P/F = 40,000/20,000 = 2m/s

Volume of water raised = 30,000 litres = 30,000/1000 m³ = 30 m³ .
Mass of water raised, M = Volume × Density
= (30m³ ) × (10³ kg/m³ )
= 30 × 10³ kg
Height, h = 45 meter
Work done by machine,
W = Weight of water raised × Height
= (Mg) × h (30 × 10³ × 9.8) × 45
W = 1.323 × 10⁷ joule
Time taken, t = 10 minutes = 10 × 60 = 600 seconds.
Power, P = W/(t ) = (1.323× 10⁷ J )/(600 s)
= 22 × 10³ W
= 22kW.

Force F = 5 ; s = 1500m + 2 × 100m + 200m
Work done, W = F.s
W= 5 × [1500 + 200 +2000]
= 18500J.

P = W/t = mgh/t
= (m×10 ×10 )/60 = 2000W
m = 12000/10
= 1200kg.

Power P = 1200/1000 = 1.2 kW
Time t, = 30/60 = 0.5h
Electric energy, E = Power × time × days
= 1.2 × 0.5 × 30
= 18kWh.

An automobile engine propels a 1000kg car along a levelled road at a speed of 36kmH-1. Find the power is the opposing frictional force is 100N, suppose after
ma = mb = 1000kg.v = 36km/h = 10m/s
Frictional force = 100N
Since that car A moves with a uniform speed it means that the engine of car applies a force equal to the frictional force.
Power = (Force × Distance)/(time) = F.v
= 100N × 10m/s = 1000W
After collision
ma ua + mb ub = ma va + mb vb
1000 × 10 + 1000 × 0 = 1000 × 10 + 1000 × vb
vb = 10 ms⁻¹

Power P = 1200/1000 = 1.2 kW
Time t, = 30/60 = 0.5h
Electric energy, E = Power × time × days
= 1.2 × 0.5 × 30
= 18kWh.

An automobile engine propels a 1000kg car along a levelled road at a speed of 36kmH-1. Find the power is the opposing frictional force is 100N, suppose after
ma = mb = 1000kg, v = 36km/h = 10m/s
Frictional force = 100N
Since that car A moves with a uniform speed it means that the engine of car applies a force equal to the frictional force.
Power = (Force × Distance)/(time) = F.v
= 100N × 10m/s = 1000W
After collision
ma ua + mb ub = ma va + mb vb
1000 × 10 + 1000 × 0 = 1000 × 10 + 1000 ×vb
vb = 10 ms⁻¹