Page No 137: - Chapter 6 Work Energy & Power class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 6.17:

The bob A of a pendulum released from 30° 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:

Bob A will not rise at all

In an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. In this case, a complete transfer of momentum takes place from the moving mass to the stationary mass.

Hence, bob A of mass m, after colliding with bob B of equal mass, will come to rest, while bob B will move with the velocity of bob A at the instant of collision.

Question 6.18:

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:

Length of the pendulum, l = 1.5 m

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, 

Total energy   … (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.,

Question 6.19:

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:

The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sandbag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, the speed of the trolley will remain 27 km/h.

Question 6.20:

A body of mass 0.5 kg travels in a straight line with velocity

 where . What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Answer:

Mass of the body, m = 0.5 kg

Velocity of the body is governed by the equation, 

Initial velocity, u (at x = 0) = 0

Final velocity v (at x = 2 m) 

Work done, W = Change in kinetic energy

Question 6.21:

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:

Area of the circle swept by the windmill = A

Velocity of the wind = v

Density of air = ρ

(a) 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

(b) Kinetic energy of air 

(c) Area of the circle swept by the windmill = A = 30 m²

Velocity of the wind = v = 36 km/h

Density of air, ρ = 1.2 kg m^–3

Electric energy produced = 25% of the wind energy

Question 6.22:

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 × 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:

(a) Mass of the weight, m = 10 kg

Height to which the person lifts the weight, h = 0.5 m

Number of times the weight is lifted, n = 1000

∴Work done against gravitational force:

(b) Energy equivalent of 1 kg of fat = 3.8 × 10⁷ J

Efficiency rate = 20%

Mechanical energy supplied by the person’s body:

Equivalent mass of fat lost by the dieter:

Question 6.23:

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:

Answer: (a) 200 m²

(a) 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

As per the information given in the Question, we have:

(A × 200)

(b) 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.

Question 6.24:

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:

Mass of the bullet, m = 0.012 kg

Initial speed of the bullet, u_b = 70 m/s

Mass of the wooden block, M = 0.4 kg

Initial speed of the wooden block, u_B = 0

Final speed of the system of the bullet and the block = ν

Applying the law of conservation of momentum:

For the system of the bullet and the wooden block:

Mass of the system, m‘ = 0.412 kg

Velocity of the system = 2.04 m/s

Height up to which the system rises = h

Applying the law of conservation of energy to this system:

Potential energy at the highest point = Kinetic energy at the lowest point

= 0.2123 m

The wooden block will rise to a height of 0.2123 m.

Heat produced = Kinetic energy of the bullet – Kinetic energy of the system

= 29.4 – 0.857 = 28.54 J

Question 6.25:

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 (Fig. 6.16). 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:

No; the stone moving down the steep plane will reach the bottom first

Yes; the stones will reach the bottom with the same speed

v_B = v_C = 14 m/s

t₁ = 2.86 s; t₂ = 1.65 s

The given situation can be shown as in the following figure:

Here, the initial height (AD) for both the stones is the same (h). Hence, both will have the same potential energy at point A.

As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i. e.,

v₁ = v₂ = v, say

Where,

m = Mass of each stone

v = Speed of each stone at points B and C

Hence, both stones will reach the bottom with the same speed, v.

For stone I:

Net force acting on this stone is given by:

For stone II:

∴

Using the first equation of motion, the time of slide can be obtained as:

For stone I:

For stone II:

∴

Hence, the stone moving down the steep plane will reach the bottom first.

The speed (v) of each stone at points B and C is given by the relation obtained from the law of conservation of energy.

The times are given as: