Question1.

The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

• Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
• work done by gravitational force in the above case,
• work done by friction on a body sliding down an inclined plane,
• work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
• Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:

Work done W = F.S

=FS cos θ    where θ= angle between F and displacement S.

• Work done is positive. In order to lift the bucket out of the well, force should be equal to weight of the bucket and has to be applied vertically upwards
• and the bucket moves along the same direction. Therefore, θ = 0⁰, therefore W = FS cos 0⁰ =FS.
• Work done is negative. The bucket moves in a direction opposite to the gravitational force which acts vertically downwards.
• The θ between gravitational force and displacement is 180⁰. Therefore,

 W = FS cos 180⁰ = - FS

• Work done is negative. Friction always acts in opposite direction to the direction of motion. Therefore θ = 180⁰.
• Work done by the sliding down in an inclined plane is (-) ive. W =FS cos 180⁰ = -FS.
• Work done is positive. Force applied to overcome friction and the displacement is in the same direction. Therefore θ = 0⁰. Work done W = FS cos0⁰ = FS
• Work done is negative. The resistive force of air acts in the direction opposite to the direction of motion of the pendulum, i.e. θ = 180⁰. Therefore W = FS cos180⁰ = - FS.

Question2.

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.

Compute the work done by the applied force in 10 s, work done by friction in 10 s,

work done by the net force on the body in 10 s, change in kinetic energy of the body in 10 s, and interpret your results.

Answer:

Given:

Mass of the body, m = 2 kg

Applied force, F = 7 N

Coefficient of kinetic friction, μ = 0.1

Initial velocity, u = 0

Time, t = 10 s

The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:

 a₁= (F/m) = (7/2) = 3.5m/s²

Frictional force is given as: f = μmg

= 0.1 × 2 × 9.8 = – 1.96 N

The acceleration produced by the frictional force:

a₂ = (-1.96)/ (2) = -0.98m/s²

Total acceleration of the body:

a = a₁ + a₂

=3.5 + (-0.98) = 2.52m/s²

The distance travelled by the body is given by the equation of motion:

s = ut + (1/2) at²

=0 + (1/2) x 2.52 x (10)² = 126m

Work done by the applied force, W_a = F × s = 7 × 126 = 882 J

Work done by the frictional force, W_f = F × s = –1.96 × 126 = –247 J

Net force = 7 + (–1.96) = 5.04 N

Work done by the net force, W_net= 5.04 ×126 = 635 J

From the first equation of motion, final velocity can be calculated as:

v = u + at

= 0 + 2.52 × 10 = 25.2 m/s

Change in kinetic energy= (1/2) mv² – (1/2) mu²

= (1/2) x 2(v² - u²)

= (2.52)² – (0)² = 635J

Question 3.

Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis.

In each case, specify the regions, if any, in which the particle cannot be found for the given energy.

Also, indicate the minimum total energy the particle must have in each case.

Think of simple physical contexts for which these potential energy shapes are relevant.

Answer:

Given:

Total energy of the system is given as:

E = P.E + K.E

Therefore, K.E = E – P.E.

Kinetic energy is positive quantity. It cannot be negative. Therefore, a particle will not exist in a region where K.E. becomes negative.

• For x > a; P.E.(V₀) > E

Therefore, K.E. becomes negative. Hence the object cannot exist in the region x>a.

• x > a and x < b; P.E.(V₀) > E

K.E. becomes negative. Hence, the object cannot exist in the region x>a.

  (c)    (x > a) and (x < b); -V₁

In the given condition regarding the positivity of K.E. is satisfied only in the region between (x>a) and (x<b).

The minimum potential energy in this case is –V₁.

Therefore, K.E. = E – (–V₁) = E +V₁. Therefore, for the positivity of the   kinetic energy, the total energy of the particle must be greater than –V₁.

So, the minimum total energy the particle must have is –V₁.

• (-b/2) < (a/2); (a/2) < x (b/2); - V₁

In the given case, the potential energy (V₀) of the particle becomes greater than the total energy (E) for (-b/2) < (a/2); (a/2) < x (b/2).

Therefore, the particle will not exist in these regions.

The minimum potential energy in this case is –V₁.

Therefore, K.E. = E – (–V₁) = E +V₁.

Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than –V₁.

So, the minimum total energy the particle must have is –V₁.

Question 4.

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx² /2,

where k is the force constant of the oscillator.

For k = 0.5 N m^–1, the graph of V(x) versus x is shown in Fig. 6.12.

Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Answer:

Given:

Total energy of the particle, E = 1 J

Force constant, k = 0.5 N m^–1

Kinetic energy of the particle, K = (1/2) mv²

According to the conservation law:

E = V + K

1 = (1/2) kx² + (1/2) mv²

At the moment of ‘turn back’, velocity (and hence K) becomes zero.

Therefore 1 = (1/2) kx²

= (1/2) x 0.5 x x² = 1

x² = 4

x= +2, x =-2

Hence, the particle turns back when it reaches x = ± 2 m.

Question 5.

Answer the following:

• The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere
• Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general.
• Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
• An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small.
• Why then does its speed increase progressively as it comes closer and closer to the earth?
• In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him.
• The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Answer:

The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket.

 According to the conservation of energy:

Total Energy (T.E.) = Potential Energy (P.E) + Kinetic Energy (K.E)

= mgh + (1/2) mv²

The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

• The gravitational force on the comet due to sun is a conservation force. Since the work done by a conservative force over a closed path is always zero (irrespective of the nature of path). Hence the work done by the gravitational force over every complete orbit of the comet is zero.
• When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height.
• Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E.
• Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.
• In second case work done is more.

Case (i):-         

Given:

Mass, m = 15 kg

Displacement, s = 2 m

Work done W = FS cosθ; where θ = angle between force and displacement

=mg cosθ

= (15 x 2 x 9.8 cos90⁰) 

=0   (because cos 90⁰ = 0)

Case (ii)

Given:

Mass, m = 15 kg

Displacement, s = 2 m

Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.

Therefore, the angle between them, θ = 0°

Work done, W = Fs cosθ

= mgs (because cos 0° = 1)

= (15 × 9.8 × 2)

 = 294 J

Hence, more work is done in the second case.

Question 6.

Underline the correct alternative:

• When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
• Work done by a body against friction always results in a loss of its kinetic/potential energy.
• The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
• In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Answer:

• Potential energy of the body decreases. A conservative force does a positive work on a body when it displaces the body in the direction of force.
• As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.
• There will be loss in the kinetic energy. Work is done against the direction of friction which reduces the velocity of the body.
• External force. Internal forces cannot produce any change in the total momentum of a body.
• Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system.
• Total linear momentum. The total linear momentum always remains conserved whether it is an elastic or inelastic collision.

Question7.

State if each of the following statements is true or false. Give reasons for your answer.

(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.

(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

(c) Work done in the motion of a body over a closed loop is zero for every force in nature.

(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. The total momentum and total energy of the system is conserved bot not of each body.
2. The external forces on the system can either increase or decrease the total energy of the system.
3. Work done in the motion of a body over a closed loop is 0 only when the body moves under the action of conservative force.
4. It is not 0 when the forces are non-conservative e.g. frictional forces.
5. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

                                                                                                                                           

Question 8.

Answer carefully, with reasons:

• In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
• Is the total linear momentum conserved during the short time of an elastic collision of two balls?
• What are the answers to (a) and (b) for an inelastic collision?
• If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Answer:

• It is because during collision (when the balls are in contact), K.E. of the balls gets converted into potential energy.
• However, K.E. before and after the elastic collision is the same.
• The total linear momentum of the system is always conserved.
• For inelastic collisions for the option (a) it is No. Because in an inelastic collision, there is always loss of kinetic energy.
• The kinetic energy before collision will always be greater than that of after collision.

For option (b) yes. The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.

• The collision is elastic. It is because the forces which are involved are conservative.

Question 9.

A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

(i) (t) ^1/2 (ii) t  (iii) (t)^3/2 (iv) (t)²

Answer:

Correct option: (ii) t

Given:

Mass of the body = m

Acceleration of the body = a

Using Newton’s second law of motion: F = ma

As both (m) and (a) are constants therefore F is also constant.

F = ma = Constant   (i)

Also acceleration a = (dv/dt) = Constant

Or dv= (Constant x dt)

=>v =at       (ii) where (a) is also constant.

This shows v ∝ t   (iii)

P = F.v

Using equations (i) and (iii), we have:

=> P∝ t

Hence, power is directly proportional to time.

Question 10.

A body is moving unidirectional under the influence of a source of constant power. Its displacement in time t is proportional to

(i) (t) ^1/2(ii) t (iii) (t) ^3/2 (iv) (t) ²

Answer:

Correct option: (iii)

Let Power = P which is constant.

Initial velocity u =0

Also, P = Fv = mav (using f = ma)

Therefore P = ma²t

=> a = √ (P)/ (mt)

Also we know that s = ut + (1/2) a t²

Thus, s = (1/2) √ (P)/ (mt) t²

=> s = (1/2) √ (P)/ (mt ^(3/2))

So s ∝ mt ^(3/2)

Question 11.

A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = (-î +2 ĵ +3 k̂) N Where î , ĵ and k̂ 

are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Answer:

Force exerted on the body, F = (-î +2 ĵ +3 k̂) N

Displacement, s = 4 k̂m

Work done, W = F.s

= (-î +2 ĵ +3 k̂). (4 k̂)

=0+0+3x4

=12J

Hence, 12 J of work is done by the force on the body.

Question 12.

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV.

Which is faster, the electron or the proton? Obtain the ratio of their speeds.

(Electron mass = 9.11 × 10^–31 kg, proton mass = 1.67 × 10^–27 kg,

1 eV = 1.60 × 10^–19 J).

Answer:

Electron is faster; Ratio of speeds is 13.54: 1

Mass of the electron, m_e = 9.11 × 10^–31 kg

Mass of the proton, m_p = 1.67 × 10^{– 27} kg

Kinetic energy of the electron, E_Ke = 10 keV = 10⁴ eV

= 10⁴ × 1.60 × 10^–19

= 1.60 × 10^–15 J

Kinetic energy of the proton, E_Kp = 100 keV = 10⁵ eV = 1.60 × 10^–14 J

For the velocity of an electron v_e, its kinetic energy is given by the relation:

E_ke = (1/2) mv_e²

Therefore, v_e = √ (2 x E_ke)/m

= √ (2 x 1.60 x 10^-15)/ (9.11 x 10^-31)

v_e = (5.93 x 10⁷) m/s

For the velocity of a proton v_p, its kinetic energy is given by the relation:

E_{Kp =} (1/2) mv_p²

Therefore, v_p = √ (2 x E_Kp)/m

v_p = √ (2 x 1.6 x 10^-11)/(1.67 x10^-27)

= (4.38 x 10⁶) m/s

Hence, the electron is moving faster than the proton.

The ratio of their speeds:

(v_e/ v_p) = ((5.93 x 10⁷)/ (4.38 x 10⁶)

=13.54:1

Question 13.

A rain drop of radius 2 mm falls from a height of 500 m above the ground.

It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height;

it attains its maximum (terminal) speed, and moves with uniform speed thereafter.

What is the work done by the gravitational force on the drop in the first and second half of its journey?

What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s^–1?

Answer:

Gravitational force on the rain drop = mg; where m = mass of the rain drop and

 g = acceleration due to gravity which is constant.

As the drop experiences constant gravitational force throughout therefore, work done by the gravitational force on each drop in each half of the journey is same.

Given:

Radius of the rain drop, r =2mm= 2 x10^-3m

Volume of the rain drop, V = (4/3) πr³

= (4/3) x 3.14 x (2 x 10^-3)³ m^-3

Density of water, ρ = 10³ kgm-3

Mass of the rain drop m = ρ V

= (4/3) x 3.14 x (2 x 10^-3)³ x 10³ kg

Gravitational force, F =mg

= (4/3) x 3.14 x (2 x 10^-3)³ x 10³ x 9.8 N

The work done by the gravitational force on the drop in the first half of its journey:

W_I = Fs

= (4/3) x 3.14 x (2 x 10^-3)³ x 10³ x 9.8 x 250

=0.082J

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i.e., W_II, = 0.082 J

As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.

Therefore, total energy at the top:

E_T = mgh + 0

= (4/3) x 3.14 x (2 x 10^-3)³ x 10³ x 9.8 x 500 x 10^-5

= 0.164 J

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.

Therefore, total energy at the ground:

E_G = (1/2) mv² +0

= (1/2) x (4/3) x 3.14 x (2 x 10^-3)³ x 10³ x 9.8 x (10)²

=1.675 x 10^-3 J

Therefore resistive force = (E_G – E_T) = -01.62J

Question 14.

A molecule in a gas container hits a horizontal wall with speed 200 m s^–1 and angle 30° with the normal, and rebounds with the same speed.

Is momentum conserved in the collision? Is the collision elastic or inelastic?

Answer:

Yes, the collision is elastic.

The momentum of gas molecule remains conserved whether collision is elastic or inelastic.

The velocity with which the gas molecule is moving = 200m/s.

When it is striking the stationary wall it is rebounding with the same velocity.

This shows that the rebound velocity of the wall remains 0.Therefore kinetic energy of the molecule remains conserved during the collision.

The given collision is an example of an elastic collision.

Question 15.

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min.

If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Answer:

Given:

Volume of the tank, V = 30 m³

Time, t = 15 min = 15 × 60 = 900 s

Height of the tank, h = 40 m

Efficiency of the pump, η = 30%

Density of water, ρ = 10³ kg/m³

Mass of water, m = (Density x Volume) = ρV = 30 × 10³ kg

Output power can be obtained as:

P_o = (Work done)/ (Time)

= (mgh)/t)

(30 x 10³ x 9.8 x 40)/ (900)

=13.067 x 10³ W

For input power P_i efficiency η is given by the relation:

η = (P_o)/ (P_i)

P_i = (13.067)/ (30) x 100 x 10³

=0.436 x 10⁵ W

P_i = 43.6KW

 Question 16.

Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball

bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision.

Answer:

Case (ii) will be possible after collision.

Given:

Mass of each ball = m

Before collision, Total K.E. of system = (1/2) mv²+ 0= (1/2) mv²

After collision, K.E. of system is;

Case (i):-

Total kinetic energy of the system after collision:

= (1/2) m x 0 + (1/2) (2m) (V/2)²

= (1/4) mV²

Hence, the kinetic energy of the system is not conserved in case (i).

Case (ii):-

Total kinetic energy of the system after collision:

= (1/2) (2m) x0+ (1/2) mv²

= (1/2) mV²

Hence, the kinetic energy of the system is conserved in case (ii).

Case (iii)

Total kinetic energy of the system after collision:

= (1/2) (3m) (V/3)²

= (1/6) m V²

Hence, the kinetic energy of the system is not conserved in case (iii).

Question 17.

The bob A of a pendulum released from 30^o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15.

How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Answer:

When two bodies of same mass undergo elastic collisions their velocities get interchanged.

In the given situation, bob A is moving with certain speed and bob B is at rest. Therefore, after collision, bob A comes to rest and the bob B starts moving with the speed of bob A.

The whole momentum of bob A will get transfer to bob B and so bob A; will not rise at all after the second collision.

Therefore bob A will come to rest and bob B will keep on moving.

 Question 18.

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m,

what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Answer:

Given:

Length of the pendulum l =1.5m

Mass of the bob = m

Energy dissipated = 5%

According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:

Potential energy of the bob, E_P = mgl

Kinetic energy of the bob, E_K = 0

Total energy = mgl ... (i)

At the lowermost point (mean position):

Potential energy of the bob, E_P = 0

Kinetic energy of the bob, E_k = (1/2) mv²

Total energy = E_T = (1/2) mv²... (ii)

As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.

The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i.e.

(1/2) mv² = (95/100) x mgl

v = √ (2x95x1.5x9.8)/ (100)

=5.285m/s

Question 19.

A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track.

After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s^–1. What is the speed of the trolley after the entire sand bag is empty?

Answer:

As the trolley carrying sand bag is moving uniformly, so the external force on the system is 0.

But when the sand leaks out, it won’t apply any external force on the trolley. As a result the speed off the trolley does not change it remains 27 km/h.

 

Question 20.

A body of mass 0.5 kg travels in a straight line with velocity v =a x (^3/2) where

a = 5 m^–(1/2) s^–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Answer:

Given:

Mass of the body, m = 0.5 kg

Velocity v = a x (^3/2)    (equation 1)

Acceleration a =5 m^–(1/2) s^–1

Initial velocity, u (at x = 0) = 0

Final velocity v (at x = 2 m)

= a2 ^(3/2) = 5 x 2^(3/2) =10√2 m/s

Work done, W = Increase in kinetic energy

= (1/2) m (v² – u²)

= (1/2) x 0.5 [(10√2)² – (0)²]

= ((1/2) x 0.5 x 10 x 10x 2)

=50J

 

Question 21.

The blades of a windmill sweep out a circle of area A.

(a) if the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air?

(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m², v = 36 km/h and the density of air is 1.2 kg m^–3.

What is the electrical power produced?

Answer:

Given:

Area of the circle swept by the windmill = A = 30 m²

Velocity of the wind = v = 36km/h

Density of air = ρ = 1.2 kg m^–3

• Volume of the wind flowing through the windmill per sec = Av

             Mass of the wind flowing through the windmill per sec = ρAv

• Mass m, of the wind flowing through the windmill in time t = ρAvt

             Kinetic energy of air = (1/2) mv²

             = (1/2) (ρAvt) v² = (1/2) ρAv³t

• Electric energy produced = 25% of the wind energy

              = (25/100) x Kinetic energy of air

              = (1/8) ρAv³t

       Electrical power = (Energy)/ (Time)

= (1/8) (ρAv³t)/ (t)

= (1/8) ρAv³

= (1/8) x 1.2 x 30 x (10)³  =4.5 x 10³ W =4.5KW

Question 22.

A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time.

Assume that the potential energy lost each time she lowers the mass is dissipated.

(a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy

with a 20% efficiency rate. How much fat will the dieter use up?

Answer:

Given:

• Mass of the weight m =10kg

Height to which weight is lifted h=0.5m

Number of times the weight is lifted, n = 1000

Work done against gravitational force: = n (mgh)

= 1000 x 10 x 9.8x0.5

=49000J.

• Energy equivalent of 1 kg of fat = 3.8 × 10⁷ J

Efficiency rate = 20%

Mechanical energy supplied by the person’s body:

= (20/100) x 3.8x10⁷ J

= (1/5) x 3.8x10⁷ J

Equivalent mass of fat lost by the dieter:

= (1/ ((1/5) x 3.8x10⁷) x 49 x 10³

= (245/3.8) x 10^-4

=6.45 x 10^-5 kg

Question 23.

A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter.

If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

(b) Compare this area to that of the roof of a typical house.

Answer:

Given:-

Power used by the family, P = 8 kW = 8 × 10³ W

Solar energy received per square metre = 200 W

Efficiency of conversion from solar to electricity energy = 20 %

Area required to generate the desired electricity = A

(8 x 10³)  = (20%) x (A x200)

= (20/100) x A x 200

Therefore A = (8x10³)/ (40)

=200m²

The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m.