CLASS 11 WORK AND ENERGY
WORK AND ENERGY
Fx = Fcosθ
Let P acting on a body & displacement occurs in Horizontally.
W = F . d Unit: Joules or Nm
W = fd cosθ Dimensions: ML²T⁻²
→ It is a scalar quantity
→ It may be +ve, -ve or zero
1) Positive Work Done:-
If θ ≤ θ < 90°
then cosθ > 0
So, w = fd is +ve
❈ When force & displacement are in same direction then work is +ve
θ = 0°
Wmax = fd
2) Zero work: -
❈ When θ = 90° the F & d are ⊥ to each other
W = fd cos 90°
W = 0
3) Negative Work :-
❈ When direction of force & displacement are opposite to each other. then work is +ve.
W= -fd
Examples:-
① work done by frictional force
② work done against gravity.
NOTE:-
Let F= f₁î+f₂ĵ+f₃k.
d=d₁î+d₂ĵ+d₃k
W = F.d
W = f₁d₁ + f₂d₂ + f₃d₃
NOTE:-
A= a₁î+a₂ĵ+a₃k, B=b₁î+b₂ĵ+b₃k
d = B - A
d = (b₁-a₁)î + (b₂-a₂)ĵ + (b₃-a₃)k
∴ W = F.d
✤ WORK DONE BY VARIABLE FORCE:-
Let f(x) is a variable then work done by this force from
x=a to x=b
W = ∫[a to b] f(x) dx
❈ POWER:
Rate of doing work by a body is called power.
Power = Work / Time
P = W / t Unit is Joules/sec or Watts.
✩ EFFICIENCY:
It to ratio of output power & input power.
η = (P₀ / Pᵢ) x 100
✤ ENERGY:
It is defined as the ability of a body to do work.
(A) Potential Energy: It is the energy possessed by a body due to its particular position.
Force due to Gravity
F = mg
: Work done to life a body of mass ‘m’ up to height ‘h’
W = f x d
W = mgh
This work is stored in the form of PE.
:. u = mgh Joules
• KINETIC ENERGY :-
It is the energy possessed by a body due to its motion let a body of mass m is moving with velocity v then KE = ½ mv² Joules
Proof :- let dw work is done in dx displacement
:. dθ = fdx
= m dv/dt.dx
= mdv (dx/dt)
∫ dw = ∫ mv.dv
w-0 = m (v²/2)ᵤᵥ [u=0]
KE = ½ mv²
✤ WORK ENERGY THEOREM :-
It states that amount of work done is always equal to the change in KE of the body
:. work = change in KE
W=Kf-Ki
Proof :- let dw work is done in dx displacement
:. dw=fdx
= m dv/dt.dx
v = mdv (dx/dt)
∫ dw = ∫ mv.dv
w-0 = m(v²/2)uv => w=½ m (v²-u²)
= ½ mv²-½ mu²
NOTE:- KE can't be negative.
❂ CONSERVATION OF ENERGY ->
It states that in freefall, sum of all energies at every point remain conserved.
Case1: Let an object of mass m is placed at 'h' height
PE=mgh
KE=0
Total Energy = mgh + 0
= mgh ①
At point B :-
PE = mg (h-x)
v² = u²+2ax
v² = 0 + 2gx
:.KE = ½ m 2gx
= mgx
Total Energy = mgn + mg(h-x)
= mgh + mgh - mgx
= mgh
At Point C:- h = 0
PE = mgh
PE = 0
Now, v² = u² + 2as
v² = 0 + 2gh
v² = 2gh
:.KE = ½ mv²
= ½ m x 2gh
Total energy = mgh
It shows that Total Energy always remain conserved.
✤ RESTORING FORCE :-
➤ It is an induced force which bring the body in its initial state of rest
i.e f ∝ -x
F = -kx
Here k is spring constant. S.I unit = N/m k = -f/x
➤ Depend upon the nature of material of spring
➤ Depend upon length & thickness.
❆ ENERGY STORED IN STRETCHED SPRING :-
Let dw work is done to displace the spring by dx against restoring force
:.dw = -fdx
:.dw = -(-kx) dx
dw = kx.dx
Integrate both sides
∫dw = k ∫xdx
W = k. x²/2
W = ½ kx²
This work is stored in the form of energy
U = ½ kx² Joules
✤ VARIATION IN ENERGY:-
E= 1/2 kx²
E ∝ x²
❆ COLLISION:-
When a body strike to another body & exert forces on it , then it is called Collision.
BEFORE AFTER
U₁ > U₂
TYPES OF COLLISION:-
① Elastic Collision:-
In this type of collision
① Momentum remain conserved of the entire system.
② K-E remain constant.
③ Total energy conserved.
Velocity of a body after collision (elastic) :-
Before After
In elastic Collision
Momentum before Collision = Momentum after collision
m₁U₁+ m₂U₂ = m₁V₁+ m₂V₂
m₁(U₁-V₁) = m₂ (V₂-U₂) ①
KE is always constant
½ m₁u² + ½ m₂u² = ½ m₁v² + ½ m₂v²
m₁(U² - V²) = m₂ (V₂² - U₂²) ②
②/①
U₁+V₁ = U₂+V₂
U₁ - U₂+V₁ = V₂
Putting in ①
m₁U₁+ m₂U₂ = m₁V₁+ m₂V₂
m₁U₁-m₁V₁ = m₂(U₁-V₂ + V₁) - m₂V₂
m₁U₁ - m₁V₁ = m₂U₁ -2m₂U₂+ m₂V₁
(m₁-m₂)U₁+ 2m₂U₂ = (m₁+m₂)V₁
V₁ = (m₁-m₂)/ m₁+m₂ . U₁+ 2m₂U₂/ m₁+m₂
V₂ = (m₂-m₁)/ m₁+m₂ . U₂ + 2m₁U₁/ m₁+m₂
❆ Inelastic Collision :-
In this collision
① Momentum remain Conserved.
② K-E does not remain Constant.
③ Total energy remain Conserved.
❈ COEFFICIENT OF RESTITUTION :-
It defined as the ratio of relative velocity after collision & relative velocity before collision.
e = V₂-V₁ / U₁-U₂
1) For elastic collision, e=1
2) For inelastic, e=0
* If m1=m2=m(let)
V1 = (m1-m2)U1/m1+m2+ 2m2U2/m1+m2
= (m-m)U1/m+m + 2mU2/2m
V1 = 0+U2
V1 = U2 & V2 = U1
For Elastic KE e=1 Between-three Collision Conserved atoms
For inelastic KE 0
Collision Non-conserved objects
Perfect No loss in e=0 In shooting elastic KE
Super Increase e>1 In explosions
Elastic in KE
✤ INELASTIC COLLISION :-
In this type of collision momentum always remain conserved & After collision two bodies moves with each other.
Momentum after Collision = m1U1 + m2U2
Let they move with Velocity V
Momentum after Collision = (m1+m2)V
By Conservation of momentum
(m1+m2)V = m1U1+ m2U2
✤ Collision in 2D :-
let two bodies of masses m1 & m2 are collided as shown in figure
Horizontal Momentum
M1u + m2x0 = m1v1cosθ1+m2v2cosθ2
M1u = m1v1cosθ1+m2v2cosθ2
& for Vertical Momentum
m1v1sin θ1 = m2v2sin θ2
Conservative force = -ve of Potential energy
* * * * * * * * * * * * * * * * * *
