Unit-6

2. Fill in the blanks with suitable words:

a. The distance covered in unit time is speed.
b. Unit of power is Watt.
c. Rate of doing work is called Power.
d. Use of ball bearings reduces friction.

3. Answer the following questions:

a. Write the unit of the following physical quantities:

Distance = Meter (m)
Displacement = Meter (m)
Work = Joule (J)
Power = Watt (W)
Speed = Meter per second (m/s)
Velocity = Meter per second (m/s)

b. Clarify distance and displacement with suitable examples:

Distance is the total path covered by an object, irrespective of the direction. It is a scalar quantity.
- Example: If a person walks 3 meters forward and then 4 meters backward, the total distance traveled is 7 meters.
Displacement is the shortest straight line between the initial and final positions of an object, along with the direction. It is a vector quantity.
- Example: If the person walks 3 meters forward and 4 meters backward, the displacement is 1 meter in the backward direction.

c. "Friction is necessary evil" — justify the statement:

Friction is considered a "necessary evil" because, it is both useful and harmful. Friction is necessary for many daily activities like walking, driving, lighting a match, writing etc. Along with this friction causes objects to wear and tear, causes loss of energy, generates heat which may cause unwanted fire.

d. Define work and write its unit:

Work is done when a force is applied to an object, causing it to move in the direction of the force applied.
- Formula: Work = Force × Displacement
- Unit: Joule (J)

e. Write any three measures of minimizing friction:

Lubrication: Applying oils or grease between moving surfaces reduces friction.
Use of ball bearings: These allow smoother motion by reducing surface contact and friction.
Polishing surfaces: Smoothening surfaces helps reduce friction by minimizing the roughness between surfaces.

f. Define power and mention its unit:

Power is the rate at which work is done.
- Formula: Power = Work / Time
- Unit: Watt (W)

g. Complete the formula of the following physical quantities:

Work = Force × Displacement
Speed = Distance / Time
Velocity = Displacement / Time
Power = Work / Time

h. "Friction also plays a supportive role for us." Justify the statement.

Friction, despite being often seen as a force that resists motion, actually plays a crucial supportive role in many daily activities. For example, when we walk, friction between our feet and the ground prevents us from slipping, making it possible for us to move. Similarly, friction between the tires of vehicles and the road surface is essential for vehicles to grip the road and move safely. Friction is also vital in other everyday tasks, such as writing. When we write with a pen or pencil, friction allows the ink or graphite to adhere to the paper, making writing possible. Thus, friction is necessary for stability and control in a variety of tasks.

i. Clarify the reason for reducing friction.

Reducing friction is important in many situations to improve efficiency and performance. For example, in machines, reducing friction means less energy is lost (in the form of heat), which makes the machine work more efficient. In vehicles, reducing friction between the tires and the road helps in reducing fuel consumption and improves speed. Additionally, reducing friction minimizes wear and tear on moving parts, which extends their lifespan. In everyday life, lubricants are used to reduce friction in things like engines, doors, and even in sports equipment, making movements smoother and faster.

J. i. What is the velocity of the car in the path AB? Why?

The velocity along the path AB is 5 m/s.

ii. What is the velocity of the car in the path BC? Why?

The velocity along the path BC is 5 m/s.

iii. What is the velocity of the car in the path CD? Why?

The velocity along the path CD is 5 m/s.

k. Is a person pushing the wall of the house doing work or not? Give reason.

No, the person is not doing work in this case. According to the physics ]work is done only when a force causes displacement in the direction of the force. If the wall doesn’t move, there is no displacement, and hence no work is done, even though the person is applying force.

l. If two persons are doing the same amount of work, in which condition their power will be the same and in which condition their power will be different? Clarify.

Power is defined as the rate at which work is done (Power = Work / Time). Even if two persons are doing the same amount of work, their power will differ depending on how long it takes to complete the work:

Power will be the same if both persons take the same amount of time to complete the work.
Power will be different if one person takes less time to complete the work, and the other takes more time. The person who takes less time will have more power.

m. If a person is walking carrying a load on his head, is he doing work or not? Clarify.

No, the person is not doing any work in this case.

Work is done when there is a displacement in the direction of the force applied. When the person walks, they are applying an upward force to carry the load, but they are displacing themselves forward and not in the direction of force applied.SO, according to physics the person is not doing any work.

Q NO. 4

a. Rohit covers a distance of 400 m by running in a straight path in 35 s. Calculate his velocity.

Given:

Distance ( $d$ ) = 400 m

Time ( $t$ ) = 35 s

Formula:
$v = \frac{d}{t}$
Calculation:
$v = \frac{400}{35} = 11.43 \, \text{m/s}$
Answer:
Velocity = 11.43 m/s

b. An object covers a distance of 100 m in 5 seconds. Calculate its speed.

Given:

Distance ( $d$ ) = 100 m

Time ( $t$ ) = 5 s

Formula:
$v = \frac{d}{t}$
Calculation:
$v = \frac{100}{5} = 20 \, \text{m/s}$
Answer:
Speed = 20 m/s

c. What distance is covered by the car in 10 seconds if it is traveling at a speed of 20 m/s?

Given:

Speed ( $v$ ) = 20 m/s

Time ( $t$ ) = 10 s

Formula:
$d = v \times t$
Calculation:
$d = 20 \times 10 = 200 \, \text{m}$
Answer:
Distance = 200 m

d. If a crane takes 20 s to lift a jeep of 1500 kg up to 120 m high, calculate the work done by the crane and its power.

Given:

Mass ( $m$ ) = 1500 kg

Height ( $h$ ) = 120 m

Time ( $t$ ) = 20 s

Gravitational acceleration ( $g$ ) = $9.8 \, \text{m/s}^2$

Solution,

Formula:
$W = F(force) \cdot h$

Calculation:

$F = m . g$ = 1500 . 9.8 = 14700 N

W = F . h = 14700 . 120 = 1764000 J

Power ( $P$ ):

Formula:
$= W / t$
Calculation:
$P = \frac{1764000 /}{20} = 88200 W$

$P = \frac{1764000}{20} = 88200 \, \text{W}$
Answer:
Work Done = 1764000 J
Power = 88200 W

e. A person carries a 50 kg rice bag at a height of 3 m and slides it 4 m applying an effort of 450 N.

i. What amount of work against gravity is done?

Given:

Mass ( $m$ ) = 50 kg

Height ( $h$ ) = 3 m

Gravitational acceleration ( $g$ ) = $9.8 \, \text{m/s}^2$

Formula:
$W = m \cdot g \cdot h$
Calculation:
$W = 50 \cdot 9.8 \cdot 3 = 1470 \, \text{J}$
Answer:
Work against gravity = 1470 J

ii. What amount of work against friction is done?

Given:

Force ( $F$ ) = 450 N

Distance ( $d$ ) = 4 m

Formula:
$W = F \cdot d$
Calculation:
$W = 450 \cdot 4 = 1800 \, \text{J}$
Answer:
Work against friction = 1800 J

f. Total distance covered by an object traveling in the path ABCD.

Given:

Speed from A to B = $4 \, \text{m/s}$ , Time = $4 \, \text{s}$

Speed from B to C = $3 \, \text{m/s}$ , Time = $2 \, \text{s}$

Speed from C to D = $2 \, \text{m/s}$ , Time = $6 \, \text{s}$

Distance Calculations:

1. Distance from A to B:
$d_{AB} = v \cdot t = 4 \cdot 4 = 16 \, \text{m}$
2. Distance from B to C:
$d_{BC} = v \cdot t = 3 \cdot 2 = 6 \, \text{m}$
3. Distance from C to D:
$d_{CD} = v \cdot t = 2 \cdot 6 = 12 \, \text{m}$
Total Distance:
$d_{\text{total}} = d_{AB} + d_{BC} + d_{CD} = 16 + 6 + 12 = 34 \, \text{m}$