**GSEB Solutions for Class 9 Science and Technology – Work, Energy and Power ****(English Medium)**

GSEB SolutionsMathsScience

**Exercise 9:**

**Solution 1.1:**

C. Statement-3

For work to be done, displacement in the direction of the applied force is necessary, which indicates that displacement takes place when force is applied.

**Solution 1.2:**

C. Reduces

As the angle between displacement and force increases, magnitude of work reduces. This happens because work (W) is given as,

Work = Magnitude of component of force in the direction of displacement x Magnitude of displacement

Or, W = FScosθ

It is clear from the above expression that as the measure of the angle increases, the value of its cos component will decrease and hence, work done shall also decrease.

**Solution 1.3:**

D. Zero

During circular motion, the displacement is always perpendicular to the centripetal force i.e. the angle between force and displacement is always 90^{o}. Hence the work done is zero.

**Solution 1.4:**

A. Kinetic energy

Kinetic energy is equal to half the product of mass and square of velocity. As mass and square of velocity can never be negative, kinetic energy cannot be negative.

**Solution 1.5:**

D. Energy

When work is done, there is an exchange in energy. Thus, Work done = Change in energy.

**Solution 1.6:**

A. 15

Given, Mass of man (m) = 60 kg

Mass of bucket (m’) = 15 kg

Total mass (M) = 60 +15 = 75 kg

Height (h) = 20 m

Acceleration due to gravity (g) = 9.8 m/s^{2
}∴ Work done = Mgh = (75) × (9.8) × (80)

or, Work done = 14700 J = 14.7 kJ ≈ 15 kJ

**Solution 1.7:**

A. 490

Work done = Force x Displacement

Work done = 49 x 10 = 490 J

**Solution 1.8:**

**Exercise 10:**

**Solution 1.9:**

**Solution 1.10:**

**Solution 1.11:**

**Solution 1.12:**

**Solution 1.13:**

A. F < F′

F’ = Fcosθ

The component of force in the direction of displacement is either equal to or less than the force itself. When theta is 180°, F’ = -F and when theta is 0°, F’ = F.

**Solution 1.14:**

C. 3.6 x 10^{6
}1 kWh = 3.6 x 10^{6 }J

**Solution 1.15:**

C. increases

When a spring is compressed, its potential energy increases.

**Solution 2.1:**

For work to be done on a body, action of force and displacement in the direction of the force is necessary.

**Solution 2.2:**

Work is a scalar quantity.

Work done is equal to the product of two vector quantities (force and displacement in the direction of force). We know that the product of two vectors results in a scalar or dot product.

**Solution 2.3:**

Work done, when force acts on a body, is equal to the product of magnitude of displacement during the time period in which force is acting, and magnitude of the component of force is in the direction of the displacement.

**Solution 2.4:**

1 Horse Power = 746 J/s

**Solution 2.5:**

One kilowatt hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for 1 hour.

It is used as the commercial unit of electricity.

**Solution 2.6:**

According to the law of conservation of energy, energy can neither be created nor destroyed. The total energy of the universe remains constant though it can be changed from one form to another.

**Solution 2.7:**

The angle between force and displacement measures zero degrees when the entire force is responsible for the work done.

**Solution 2.8:**

We know that,

E = mc^{2
}Here, m = 1 kg and c = speed of light = 3 × 10^{8}m/s

∴ E = (1 kg)(3×10^{8}m/s)^{ 2
}∴E= 9 ×10^{16}joule

**Solution 2.9:**

For a given system, the sum of kinetic energy and potential energy is known as mechanical energy.

**Solution 2.10:**

In stars, energy is produced by the process of nuclear fusion.

In this process, lighter nuclei like hydrogen, deuteron etc. combine with each other at high temperatures and form a helium nucleus. In this reaction, large quantity of nuclear energy is released.

**Solution 2.11:**

The main use of pulleys is to change the direction of application of force to a convenient direction i.e. the downward direction.

**Solution 3.1:**

Work is said to be done when the force applied on a body makes the body move (i.e., there is displacement of the body).

For example, a man pushing a car and a cyclist while pedaling a cycle do work, but a man who tries to push a wall, does not do any work as the wall does not move.

**Solution 3.2:**

**Solution 3.3:**

The ability of a body to do work due to its position or configuration is known as the potential energy of the body.

As shown in the figure below, let the potential energy at the reference level be zero.

Let us apply a force, having magnitude equal to the gravitational force acting on it, in the opposite direction. It is moved with constant velocity to a height h.

Work done = Force x Displacement

As the magnitude of applied force is equal to the magnitude of gravitational force,

Force = mg x h

Energy spent in doing this work is stored in the body in the form of potential energy, so potential energy at a height h from the reference level can be given by the following formula.

U = mgh

**Solution 3.4:**

Unit of electrical energy consumed is 1 kWh.

One kilowatt hour is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.

Thus,

1 kWh = 1 kilowatt x 1 hour

= 10^{3} watt x 3600 s

= 10^{3} J/s x 3600 s

= 3.6 x 10^{6} J

Thus, 1 kWh = 3.6 x 10^{6} J

**Solution 3.5:**

The energy released during the processes of nuclear fission and fusion is called nuclear energy. In both these processes, there is a loss in mass which converts into energy in accordance with the Einstein’s mass-energy relation, E = mc^{2}. This energy is used in nuclear reactors to produce electrical energy. Energy is produced in the Sun and other stars by nuclear fusion.

**Exercise 11:**

**Solution 4.1:**

**Note: **The answer given in the text book is incorrect.

**Solution 4.2:**

Given,

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 20 s

We know that,

Work done = mgh

or, W = 10 × 10 × 8 = 800 J

**Solution 4.3:**

**Solution 4.4:**

**Solution 4.5:**

Power of one tube light = 40 W

Power of 4 tube lights = 4 × 40 = 160 W = 0.16 kW

Energy consumed by 4 tube lights each day = 0.16 × 5 = 0.8 kWh

Power of one fan = 120 W

Power of 3 fans = 3 × 120 = 360 W = 0.36 kW

Energy consumed by 3 fans each day = 0.36 × 4 = 1.44 kWh

Total electrical energy consumed per day = 0.8 + 1.44 = 2.24 kWh

∴total number of units consumed per day = 2.24 units

Cost of one unit =5

Number of days in the month of September = 30 days

∴ Electricity bill for the month of September = 2.24 × 30 × 5 =336