Energy : The Driving Force – Maharashtra Board Class 9 Solutions for Science and Technology

Energy : The Driving Force – Maharashtra Board Class 9 Solutions for Science and Technology  (English Medium)

AlgebraGeometryScience and TechnologyHindi

Solution 1:

1. The energy stored in the dry cell is in the form of chemical energy.
2. The work done is zero if there is no displacement.
3. Flowing water has kinetic energy.
4. By stretching the rubber strings of a catapult we store potential energy in it.

Solution 2.1:

Work is the product of the force acting on the body and the displacement of the body.
W = Fs

Solution 2.2:

Energy is the capacity to do work.

Solution 2.3:

Power is rate of doing work.

Solution 2.4:

The energy possessed by a body by virtue of its motion is called kinetic energy.

Solution 2.5:

The energy possessed by a body because of its shape or position or configuration is called potential energy.
PE = mgh

Solution 3.1:

1. Dams are constructed on the river to store a large quantity of water at a great height so that it has potential energy.
2. This potential energy can be converted into kinetic energy by allowing water to flow under gravity to a lower level.
3. The kinetic energy can be converted into electrical energy using turbines and generators. Therefore, dams are constructed on the river.

Solution 3.2:

1. In a river, water is flowing which exerts a force on the swimmer. The swimmer has to overcome this resistive force and hence requires a lot of energy.
2. While in a swimming pool, there is no flow of water, so the swimmer does not have to overcome the resistive force. Therefore, it is easy to swim in a swimming pool than in a river.

Solution 4:

In this case, the displacement of a body is in the direction of force (F) applied on the body as shown in the figure

So, the work done by the force is­ given by
W = Fs

Solution 5:

Let ‘m’ be the mass of a moving body.
Let ‘u’ be the initial velocity, ‘v’ be the final velocity and ‘a’ be the acceleration of the body.
Using the equation of motion, we get

Solution 6:

KE (J)	m(kg)	v (m/s)
3150	28	15
2000	10	20
9520	1190	4

Solution 7.1:

Data: M = 17 kg, g = 9.8m/s², s = 2 m, W = ?
Force applied = mg
= 17 kg × (9.8 m/s²)
= 166.6 N (downward)
Force applied by Ajit to lift the load
= 166.6 N (upward)
W = Fs
= 166.6 × 2 m = 333.2 J
The work done by Ajit is 333.2 J.

Solution 7.2: