WORK, ENERGY AND POWER
IMPORTANT QUESTIONS OF PAST PAPERS:
1.Define elasticity, work and energy. (2008)2.Define kinetic energy and potential energy and write difference between them. (2009), (2013), (2007)
3.Define work and write down its formula. (2011)
4.Define work and energy state law of conservation of energy. (2008), (2013)
5.Define kinetic energy and potential energy derive the equation K.E = 𝟏/2mv2 (2012), (2014), (2009).
6.Define power and derive equation P=FV. ( 2014), (2016)
7.Define interconversion of kinetic energy and potential energy. (2015)
------------
IMPORTANT SYMBOLS:
• Applied force = F
• Angle = ɵ
• Distance = d
• Weight = W
• Speed = V
• Work = w
• Mass = m
• Power = p
• Height = h
• Kinetic energy = K.E
• Gravity = g
• Potential energy = P.E
-------------
IMPORTANT FORMULA’S:
• W= f×d• W= fdcosɵ
• K.E = 𝟏/2 mv2
• P = w/ t
• P = FV
• W =F= mg
• P.E = mgh
• Gain in K.E = Loss in P.E
𝟏/2mv2 = mgh
--------------
Questions/ Answer
Q.1: Define Work, Power, Energy and Elasticity with formula and units also.Ans:WORK: When a force acts on a body and body cover some displacement along the direction of force, then work is said to be done.
Work depend upon two physical quantities Force and displacement therefore work can be define as,
“The product of force and displacement is equal to work.”
Formula: Work = Force × Displacement
W = F × d
Unit: In S.I system the unit of work is joule and it is denoted by j.
•If ɵ = 0° when work is positive
•If ɵ = 180° when work is negative
•If ɵ = 90° when work is zero.
-------------
POWER:“The work done by a body in unit time is called power.”
OR
“The rate of doing work is called work.”
“The rate of doing work is called work.”
Formula:
Unit: The unit of power is j / s .
----------------
ENERGY: The capability of doing work is called energy. Energy is an agent which is responsible to do work.
Unit: its unit is joule.
----------------
ELASTICITY: It is ability of a body to resist a distorting influence and to return to its original size and shape when that influence or force is removed.
----------------
Q.2: Define kinetic energy and potential energy.
Ans:KINETIC ENERGY:& The energy possess by body by virtue of its motion is called kinetic energy. It value increases with velocity.
Formula: It can be calculated by K.E =& 𝟏/2 mv2
POTENTIAL ENERGY: The energy possessed by a body by virtue of position is called potential energy. It value increases with increase in altitude.
Formula: It can be calculate P.E= mgh.
--------------
Q.3: Define power and derive equation P=FV.
Ans:POWER: “The work done by a body in unit time is called power.”
Suppose a constant force “F” acts on a body and displaces it through distance “d” in the direction of force in time “t”, the work is done,
W = Fd
Average power is developed,
Put the value of d / tin equation (ii)
P = FV
--------------
Q.4: Define potential energy and also derive P.E = mgh.
Ans: POTENTIAL ENERGY: The energy possessed by a body by virtue of position is called potential energy.
Mathematically: Consider a body of mass “m” on the ground. If it is lifted up with a vertical height “h” the force required to raise the body is just equal and opposite to its weight
W = mg.
thus work done on it against gravitational field is store in its gravitational work,
W = F.d
But, F = W = mg
W = mg.d (d = h)
W = mgh (W = P.E)
P.E = mgh
But, F = W = mg
W = mg.d (d = h)
W = mgh (W = P.E)
P.E = mgh
-------------
Q.5: Prove that Elastic potential energy = 𝟏/2 kx2
Ans:ELASTIC POTENTIAL ENERGY: The energy stored in a stretched or compressed elastic material like (spring or rubber band) is called Elastic potential energy.
Derivation: Consider an object is attached with a spring having mass “m” placed on a smooth and horizontal surface. A force f pushes a spring to compress it from its equilibrium position to another position.
According to Hooks Law, the applied force is directly proportional to the extension produce in the spring of distance “x”, so mathematically.
F œ x
F=Kx
Where “K” is constant (spring constant) its value depends on the stiffness of the spring, since the compression is zero at “O” and Kx at x, the average force needed to compress the spring from position “O” to x isF=Kx
This work causes the elastic potential energy, therefore elastic potential energy = 𝟏/2 kx2
--------------
Q.6: What do you mean by conservation of energy? State law of conservation of energy.
Ans: CONSERVATION OF ENERGY: Conservation of energy means that the amount of energy in the universe is fixed. Although energy can change forms or we can say that one kind of energy can change into another kind of energy. However the amount of energy in the universe remains the same.
LAW OF CONSERVATION OF ENERGY: Energy can neither be created nor be destroyed but it can be changed from one form to another form.
-----------
Q.7: What happens to the potential energy of a body when dropped from certain height?
Ans: When a body is dropped from a certain height, it starts losing it P.E due to downward motion. By the time it reaches the ground the whole P.E is stores as energy.
-----------
Q.8: Define kinetic energy derive the equationK.E = 𝟏/2mv2?
Ans: KINETIC ENERGY:
The energy possess by body by virtue of its motion is called kinetic energy.
MATHEMATICALLY: Consider a body mass “m” is initially at rest (Vi = 0). Now force F is applied on the body and body acquire velocity V (Vf = V).
If body covers some displacement “S” then work done is given by,
W = F.S……………………………… (i)
If force “F” produces acceleration “a” then according to Newton’s law
F = ma ……………………………..… (ii)
In order to find the value S we use third equation of motion,
2as= Vf2 – Vi2
2as = V2 – 0
2as = V2 – 0
-----------
Q. 9: Define interconversion of kinetic energy and potential energy.
Ans: INTERCONVERSION OF K.E AND P.E: Interconversion of K.E can be converted into P.E and P.E can be converted into K.E.
EXPLANATION: Consider a body of mass “m” lying at height “h” from the ground this position it has P.E = mgh and K.E = 0. Now the body is allowed to fall under the action of gravity.
As the body move downward closer to the ground its P.E keeps on decreasing and its K.E keep on increasing. It is due to the fact that h is decreasing while v is increasing. When the body just hits the surface of the ground its P.E is zero and K.E is maximum,
Gain in K.E = Loss in P.E P.E= mgh
1 / 2 mv2 = mgh
1 / 2 v2 = gh
V2 = 2gh
V = √2gh
---------------
Prepared by: Sir Waseem
No comments:
Post a Comment