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

AlgebraGeometryScience and TechnologyHindi

**Solution 1:**

- The energy stored in the dry cell is in the form of
**chemical energy**. - The work done is zero if there is no
**displacement**. - Flowing water has
**kinetic**energy. - 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:**

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

**Solution 3.2:**

- 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.
- 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/s^{2}, s = 2 m, W = ?

Force applied = mg

= 17 kg × (9.8 m/s^{2})

= 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:**