Unit 1

Q.1) how much net force is required to accelerate a 1000 kg car at 4.00 m/s2?

Solution:

Given,

A = 4.00 m/s2

M = 1000 kg

Therefore,

F = ma

= 1000 × 4

= 4000 N

Q.2) Aimee has a toy car of mass 2 kg. How much force should she apply on the car so that it should travel with the acceleration of 8 m/s2?

Solution:

Known,

M (Mass of toy car) = 2 Kg,

A (Acceleration) = 8m/s2,

F is Force to be applied by Aimee = m × a

= 2 Kg × 8 m/s2 = 16 Kgm/s2 = 16 N.

Q.3) A hammer having a mass of 1 kg going with a speed of 6 m/s hits a wall and comes to rest in 0.1 sec. Compute the obstacle force that makes the hammer stop.

Solution:

Given,

Mass of Hammer, m = 1 kg

Initial Velocity, u = 6 m/s

Final Velocity, v = 0 m/s

Time Taken, t = 0.1 s

The acceleration is: a = (v – u)/t

Therefore, a = -60 m/s2 [-ve sign indicates retardation]

Thus, the retarding Force, F = ma = 1 × 60 = 60 N

Q 4) what is the Line of Action of a Force?

Answer 4) the line along which a force is acting on an object is called the line of action of the force. The point where the force is acting on an object is called the point of application of the force. The force which opposes the relative motion between the surfaces of two objects in contact and acts along the surfaces is called the force of friction. Galileo experimentally proved that objects that are in motion move with constant speed when there is no force acting on it. He could note that when a sphere is rolling down an inclined plane, its speed increases because of the gravitational pull which is acting on it.

When all the forces acting on an object are balanced, the net force acting is zero. But, if all the forces acting on a body result in an unbalanced force, then the unbalanced force can accelerate the body, which means that a net force acting on a body can either change the magnitude of its velocity or change the direction of its velocity. For example, when many forces act on a body, and the body is found to be at rest, we can conclude that the net force acting on the body is zero

Q 5) A 60-watt bulb is switched on 24 hours a day and there is another 60-watt bulb that is turned on for only 12 hours. Find the energy consumed by both the bulbs in one day.

Solution:

For the first 12 hours, both bulb A and B are turned ON, therefore,

Power = 60 + 60 = 120 watts

Energy = Power x Time

= 120 x 12

= 1.44 kWh (kilowatt-hour)

Now for the next 12 hours only bulb A would remain ON hence,

Power = 60 watts

Energy = 60 x 12 = 0.72 kW h

In this scenario, the power consumed during the whole day varies as one bulb is turned ON for only 12 hours, so we have to calculate average power,

Average Power = Total energy consumed / Total time taken Therefore, the average power for our light bulbs will be,

= (1.44 + 0.72) / 24

= 0.092 kW

Q 6) Determine the temperature if 200 J of heat is released by the body of mass 6 Kg and has a specific heat of 0.8 J/KgoC.

Solution:

Given:

Heat released Q = 200 J,

Mass m = 6 Kg,

Specific Heat c = 0.8 J/KgoC

The temperature is given by ΔT = Q / mc

= 200 / 6 x 0.8

ΔT= 41.66oC .

Q 7) Determine the heat released when the temperature changes by 40oC by a body of mass 3 Kg which has a specific heat of 0.7 J/kgoC

Solution:

Given:

Temperature change = 40oC ,

Mass m = 3 kg,

Specific heat c = 0.7 J/kgoC ,

The Heat released is given by formula Q = mc ΔT

= 3 × 0.7 × 40

Q = 84 J

Q. 8) Explain Internal Energy:

Answer 8) as usual, we get change in internal energy = U2 – U1 = m x Cv (T2 – T1).

Another way of determination of change in internal energy is very common in isentropic operation.

By first law of thermodynamics as applied to non-flow process,

Heat supplied = change in internal energy + work done; but heat supplied is zero

Change in internal energy = – work done.

Thus, we get an important relation in an isentropic process. This relation can be stated as “Change in internal energy is numerically equal to work done”. When the work is done by the gas, it loses internal energy and it gains internal energy when the work is done on the gas.

Q.9) Explain process?

Answer 9) Process:

Thermodynamics process represents a transition in which a system changes from one state to another. When the path is completely specified then the change of state is called a process. A Process is defined as the transformation of the system from one fixed state to another fixed state .When any one of the properties changes, the working substance or system is said to have undergone a process.

Some of the processes are identified by special names as given below:

- Isobaric process (constant pressure process)
- Isochoric process (constant volume process)
- Isothermal process (constant temperature process)
- Isentropic process (constant entropy process)
- Adiabatic process(perfectly insulated process)

Q. 10) Explain temperature?

Answer 10) Temperature is the degree or intensity of the heat present in a substance or a system, expressed based on the comparative scale and shown by a thermometer.

In other words, Temperature is the measure of hotness or coldness of a body measured using Celsius, Kelvin, and Fahrenheit scales.

The change in temperature is based on the amount of heat released or absorbed. The S.I unit of temperature is Kelvin.

The Temperature formula is given by,

Δ T = Q / mc

Where,

Δ T = temperature difference,

Q = amount of heat absorbed or released,

m = mass of the body,

c = specific heat of the body.