**WORK DONE**

(i) By Constant force

W = (f cosq) S

Work done by constant force = (force). (Displacement in direction of force

= (displacement). (force in direction of displacement)

W=F.S

W = FS Cos

Here W is Scalar but it may be **+**, ─, and zero

Work done will be equal to zero if

(i) Cosq = 0 = q = 90^{o} = F ^ S or F ^ V

Ex. Work done by centripetal or centrifugal force is always zero.

Fm=q (V×B)

Fm⊥V, Fm⊥B

(2) Displacement = 0, S = 0 in circular motion

**Negative work done **

If cosq → negative, 180^{o} ³ q ³ 90^{o}

W → ─

Note: Work done by frictional force may be negative.

**Example**:- It is given that a particle is moving from initial position (2,3,5) to final position (4,1,2) under the action of constant force F=2i+3jN. what will be the work done by the constant force.

**S��lution**-Initial position r1=2i+3j+5K as given (2,3,5)

Final position r2=4i+j+2k as given (4,1,2)

Now S=r2-r1=2i-2j-3k

F=2i+3j

Now W=F.S.

=2i+3j. (2i-2j-3k)

= 4 -6 + 0.

W = -2j **Ans**

(ii) Work done by variable force

**Case I: Applying integration force**

W=F.dx=F.dy=F.ds

or W= F.ds or W=F.ds

**Example**:- if a variable force F depends on displacement X as F = (3x^{2} -5) due to this force body displaces from x = 2 mater to x = 5 meter find the work done by this variable force.

**Solution**: Given F = 3x^{2} – 5, displacement from x = 2 meter to x = 5 meter

W=25fdx=253x2-5dx

=3 25x2dx-525dx

3x3325-5X25

*=**3**3**×**5**3**-**2**3**-5**5-2*

= [125 – 8] – 5 x 3

W = 102 J

**Case II: Applying graphical method. **

W = F.S if we have group F vs S then area under the S Curve will give us work done.

Work done = Area under the F – S curve

Area = 1/2 x basic height + length x width

= 1/2 x S x F + F x S

= 3/2 FS

W = 3/2 FS

**Example**:- Spring force F = - K*x* {Here K is Force constant} it a spring elongate by 5cm and force constant K is 100N/meter then find work done on stretching the spring.

**Solution:**

Now F = -K *x*

= -100x

W=Fdx

=-10000.05xdx

*-100**x**2**2**0**0.05*

*-**100**2**0.05**2**=-50×25×**10**-4*

= 1205 x 10^{-14}

W = - 0.125 J **Ans.**

**Question:*** *An ideal massless spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless inclined at 30° to the horizontal. A 10 kg block M is released from rest at the top of the incline and brought to rest momentarily after compressing the spring by 2 m. If g = 10 ms^{-2}, what is the speed of mass just before it leave the spring?

a 20 ms-1 b 30 ms-1

c10 ms-1 d 40 ms-1

**Solution: **F = *kx*

∴ k=Fx=1001

= 100 N/m

E_{i} = E_{f}

∴12×10×v2=12×100×22-1010(2sin30°)

Solving we get,

v=20 m/s **Answer (a)**

**KINETIC ENERGY **

Scalar quantity, always will be positive K = ½ mv^{2}

**Work energy theorem: **It is given that a mass is having initial velocity u accelerate and final velocity V and covered a distance S. Here v > u

Initial kinetic energy K1=12mu2

Final kinetic enrgy=K2=12mv2

**Using equation of motion**

v^{2} = u^{2} + 2a s

12m v2-u2=2as12m

12mv2-12mu2=maS

K_{2}-K_{1} = FS

ΔK = W

**Question***: *A body of mass m = 10^{-2} kg is moving in a medium and experiences a frictional force F = -kv^{2}. Its initial speed is v_{0} = 10 ms^{-1}. If after 10s, its energy is *1**8**m**v**0**2**,* the value of k will be **(2017 JEE Main)**

(a) 10^{-3} kg^{-1} (b) 10^{-4} kgm^{-1}

(c) 10^{-1} kgm^{-1}s^{-1} (d) 10^{-3} kgm^{-1}

**Solution:** Given, force, F=kv2

∴Acceleration,a= -kmv2

or dvdt= -kmv2

⇒ dvv2= -km∙dt

Now, with limits, we have

10vdvv2=-km0tdt

⇒ -1v10v=-kmt

⇒ 1v=0.1+ktm

⇒ v=10.1+ktm=10.1+1000k

⇒ 12×m×v2=18×v02

⇒v=v02=5

⇒ 10.1+1000k=5

Þ 1 = 0.5 + 5000 k

⇒ k=0.55000⇒k=10-4kg/m **Answer (b)**

Potential Energy: Scalar quantity may be +, - , Zero

Here F= -dUdrHere U is potential energy

F.dr=-dU

**Note:** if we find slope (tanq) of u –r graph then we will find amount of force.

**BY THE U –R GRAPH **

Slope at point A is positive

Slope at point B is zero here force at B will be zero similarly slope at C is negative slope at point D is zero.

**TYPES OF POTENTIAL ENERGY**

**(1) Elastic potential energy**: Energy associated with state of Compression or expansion of an elastic object like spring. U = ½ k*x*^{2 }Here *x* is the stretch of compression.

**(2) Electric Potential energy**: Energy associated with state of separation between charged particle

**(3) Gravitational potential energy**

U=-Gm1m2r

**Note**: Attraction force the An U → negative always

U=mgh1+h/r

If h <<<<< R

U = mgh

**Example:** A chain is held on a frictionless table with (1/n)th of its length hanging over the edge if the chain has a length L and a mass M. how much work is required to pull the hanging part back on the table.

amgl8n2 bmgl4n2

cmgl2n2 dmgln2

**Solution:** L length chain mass –M

I length chain mass –M/L

dy length chain mass MLdy

W = DU

W=Mgl2n2

du_{1} = (dm) gy

du1=MLg0l/nydy

U1=mLgY220Ln=MLgL22n2

U1=MgL2n2,

U_{2} = 0 at table

**CONSERVATIVE AND NON-CONSERVATIVE FORCE **

**Conservative forces**: Work done by conservative forces are independent of path. It only depends on initial and final position.

Ex:- Gravitational forces, electrostatic forces etc.

Note: Work done by conservative forces in closed path is always zero.

W_{path1} = W _{path2} = W _{path3 }

Non conservative force: Work done by non-conservative forces depends on path.

**Ex:-** frictional force, viscous force etc.

Power Work done per unit time P =F.V

As we know W=∆K=∆U

P=dwdt=w∆t=dkdt=dudt

Unit. J/sec or watt

1 horse power = 746 watt.

**CONSTANT POWER **

Case I:- dependency of v on t.

Let an car of mass m accelerates staring from rest, while the engine supplies constant power P then how the V depends on t.

As we know P = Fv

P=mvdvdt

Pdt=mvdv

pt=mv22=V2=2ptm

V=2Ptm=V∝t1/2

**Case II:- dependency of S on t **

Let above result V=2ptm `

dsdt=2Pmt

*0**s*ds=2pmt dt

S=2Pmt3/23/2

S= 8P9m1/2 t3/2

S∝t3/2

**Question:** A uniform chain of length pr lies inside a smooth semicircular tube AB of radius r. assuming a slight disturbance to start the chain in motion, the velocity with which it will emerge from the end B of the tube will be

agr1+2π b 2gr2π+π2

c gr π+2 d πgr

**Solution:** dm=mπ dθ

∴ dU=dmgh= mπdθgrsinθ

Ui= 0πdU=2mgrπ

Now, Ki + Ui = Kf + Uf

∴ 0+2mgrπ=12mv2-mgπr2

v=2gr 2π+π2 ** Answer (b)**

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