Work Energy Power

NCERT Solution

Part 3

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?

Answer: Volume of air passing per second = A × v

Mass of air passing per second = A × v × ρ

Mass of air passing in t second = A × v × ρ × t

(b) What is the kinetic energy of the air?

Answer: Kinetic energy `=1/2mv^2`

`=1/2A×v×ρ×t×v^2`

`=1/2A×v^3×ρ×t`

(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: v = 36 km/h `=36xx5/(18)=10` m/s

Kinetic energy `=1/2A×v^3×ρ`

`=1/2×30×10^3×1.2=1800` J

25% of 18000 `=18000xx(25)/(100)=4500` J

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?

Answer: `W=mg\h`

`=10xx9.8xx0.5=49` J

Work done in lifting the weight 1000 times = 49000 J

(b) Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer: If 4900 is 20% of energy produced by fat then total energy produced by fat

`=49000xx(100)/(20)=245000` J

Amount of fat burnt `=(245000)/(3.8xx10^7)`

`=(245000xx10^(-6))/(38)=6.45xx10^(-3)` 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?

Answer: If 8 kW is 20% then total incident solar energy

`=8xx(100)/(20)=40` kW

So, area required `=(40000)/(200)=200` square meter

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

Answer: Roof of reasonably large house is same as this area.

Additional Exercises

Question 24: A bullet of mass 0.012 kg and horizontal speed 70 m s^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer: m₁ = 0.012 kg, m₂ = 0.4 kg, u₁ = 70 m/s and u₂ = ?

When the bullet goes into the block both behave as a single body

As per conservation of momentum theory

`m_1u_1=(m_1+m_2)u_2`

Or, `0.012xx70=(0.012+0.4)u_2`

Or, `u_2=(0.84)/(0.412)=2.04` m/s

Now, PE of combination = KE of combination

Or, `(m_1+m_2)gh=1/2(m_1+m_2)v^2`

Or, `0.412xx9.8xx\h=1/2xx0.412xx2.04^2`

Or, `h=(2.04xx2.04)/(2xx9.8)=0.212` m

Amount of heat produced = amount of heat lost

W= Initial KE of bullet – Final KE of combination

KE of bullet `=1/2mv^2=1/2xx0.012xx70^2`

`=0.006xx4900=29.4` J

KE of combination `1/2xx0.412xx2.04^2`

`=0.412xx1.02xx2.04=0.86` J

So, W = 29.4 – 0.86 = 28.54 J

Heat `=W/J=(28.54)/(4.2)=6.8` Calorie

Question 25: Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track. Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60° and h = 10 m, what are the speeds and times taken by the two stones?

Answer: We know, `1/2mv^2=mg\h`

Or, `v^2=2gh`

Or, `v=sqrt(2gh)`

`=sqrt(2xx10xx10)=10sqrt2=14.14` m/s

Length of AB can be calculated as follows:

Sin 30° `=h/(AB)`

Or, `1/2=(10)/(AB)`

Or, AB = 20 m

Time `=(20)/(14.14)=1.41` s

Length of AC can be calculated as follows:

Sin 60° `=h/(AC)`

Or, `(sqrt3)/2=(10)/(AC)`

Or, AC `=20/(sqrt3)=(20)/(1.732)=11.55`

Time `=(11.55)/(14.14)=0.82` s

Question 26: A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^-1 as shown in this figure. The block is released from rest with the spring in the un-stretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Answer: Given, m = 1kg,k = 100 N/m and x = 0.1 m

At equilibrium

Normal reaction, R = mg Cos 37°

Frictional force f = μR = mg Sin 37°

Net force acting on block

= mg Sin 37° - f

= mg Sin 37° - μmg Cos 37°

= mg(Sin 37° - μ Cos 37°)

At equilibrium, work done by block is equal to potential energy of spring

So, mg(Sin 37° - μ Cos 37°) = `1/2kx^2`

Or, 1 × 9.8 (Sin 37° - μ Cos 37°) = `1/2xx100xx0.1`

Or, Sin 37° - μ Cos 37° = `5/(9.8)`

Or, 0.602 - μ × 0.799 = 0.510

Or, μ × 0.799 = 0.602 – 0.510 = 0.092

Or, `μ=(0.092)/(0.799)=0.115`

Question 27: A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 m s^-1. It hits the floor of the elevator (length of elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Answer: Potential energy of bolt = mgh

= 0.3 × 9.8 × 3 = 8.82 J

Relative velocity of bolt with respect to elevator is zero. So, lost heat is equal to potential energy of bolt.

In case the lift being stationary the answer would be same as relative velocity of bolt with respect to elevator will be zero.

Question 28: A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^-1 relative to the trolley in a direction opposite to its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Answer: Given, m₁ = 200 kg, m₂ 20 kg, v₁ = 36 km/h = 10 m/s,

Initial momentum of the system = (200 + 20) × 10 = 2200 kg m/s

If v₂ is the final velocity of trolley then relative velocity of boy relative to ground

= v₂ - 4

Final momentum = m₁v₂ + m₂(v₂ - 4)

`=200v_2+20v_2-80`

Or, `220v_2-80`

According to law of conservation of momentum

Initial momentum = final momentum

Or, `2200 = 220v_2-80`

Or, `220v_2=2280`

Or, `v_2=(2280)/(220)=10.36` m/s

Now, length of trolley = 10 m

So, time taken by boy to cross this distance = `(10)/4=2.5` s

Distance covered by trolley = 10.36 × 2.5 = 25.9 m

Question 29: Which of the following potential energy curves cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between the centers of the balls.

Answer: Figure (v) describes the elastic collision of two billiard balls.

We know that potential energy varies inversely with distance

`V(r)∝1/r`

Where, r is the distance between two balls. When the two balls touch each other distance between them is double of their radii, i.e. r = 2R where R is the radius. At this point, their potential energy will become zero.

Figure (i) shows a straight line followed by sudden vertical dip. Figure (ii) shows increase in PE. Figures (iii) and (iv) are almost similar and are showing increase followed by decrease in PE. In figure (vi) PE never becomes zero. So these figures do not describe the elastic collision between two balls.

Question 30: Consider the decay of a free neutron at rest: n → p + e^-. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus.

(Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e^-, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. the correct decay process of neutron is: n → p + e^-1 + v)

Answer: The decay process of free neutron at rest is given by following relation:

`n→p+e^-`

According to Einstein's mass-energy relation, the energy of electron is given as follow:

Δmc²

Here, Δm = mass defect

= Mass of neutron – (Mass of proton + Mass of electron)

And c = speed of light

Since masses of electron, proton and neutron are constant so Δm is constant.

So, the given two-body decay is unable to explain the continuous energy distribution in β-decay of a neutron or a nucleus. Thus, the presence of neutrino on the LHS of the decay equation explains the continuous energy distribution.