## Work done,Kinetic energy and Conservative and non-conservative force

**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)**

## Equilibrium

** (1) STABLE EQUILIBRIUM**

**(2) UNSTABLE EQUILIBRIUM **

**(3) NEUTRAL EQUILIBRIUM**

**Question***: *The given plot shows the variation of U, the potential energy of interaction between two particles with the distance separating them r.

1. B and D are equilibrium points

2. C is a point of stable equilibrium

3. The force of interaction between the two particles is attractive between points C and D and repulsive between D and E

4. The forces of interaction between particles is repulsive between

points E and F.

Which of the above statements are correct?

(a) 1 and 2 (b) 1 and 4

(c) 2 and 4 (d) 2 and 3

**Solution:** At C, potential energy is minimum. So, it is stable equilibrium position.

Further,

Negative force means attraction and positive force means repulsion.

**Question***: *The potential energy f in joule of particle of mass 1 kg moving in *x*-y plane obeys the law, f = 3*x *+ 4y. Here,* x* and y are in metres. If the particle is at rest at (6m, 8m) at time 0, then the work done by conservations force on the particle from the initial position to the instant when it crosses the *x*-axis is

(a) 25 J (b) -25 J

(c) 50 J (d) -50 J

**Solution:**j

Since, particle was initially at rest. So, it will move in the direction of force.

We can see that initial velocity is in the direction of PO. So the particle will cross the X-axis at origin.

K* _{i}* + U

*= K*

_{i}*+ U*

_{f}

_{f} 0 + ( 3 x 6 + 4 x 8) = K* _{f}* + (3 x 0+ 4 x 0)

or K* _{f}* = 50 J

**Answer (c)**