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 = 90o = 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, 180o ³ q ³ 90o

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 = (3x2 -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 = 3x2 – 5,     displacement from x = 2 meter to x = 5 meter

W=25fdx=253x2-5dx =3 25x2dx-525dx 3x3325-5X25 =33×53-23-55-2 = [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 = - Kx {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 -100x2200.05 -10020.052=-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

Ei = Ef

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 mv2 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

v2 = u2 + 2a s

12m v2-u2=2as12m 12mv2-12mu2=maS K2-K1 = FS

ΔK = W

Question: A body of mass m = 10-2 kg is moving in a medium and experiences a frictional force F = -kv2. Its initial speed is v0 = 10 ms-1. If after 10s, its energy is 18mv02, the value of k will be                                                                                                               (2017 JEE Main)

(a) 10-3 kg-1                                               (b) 10-4 kgm-1

(c) 10-1 kgm-1s-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 kxHere 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 du1 = (dm) gy

du1=MLg0l/nydy U1=mLgY220Ln=MLgL22n2 U1=MgL2n2, U2 = 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.

Wpath1 = 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 0sds=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= mπgrsinθ Ui0πdU=2mgrπ Now,  Ki + Ui = Kf + Uf

∴     0+2mgrπ=12mv2-mgπr2 v=2gr 2π+π2 Answer (b)

Join any of the batches using this book

## Batch List #### JEE 2023 Physics Online Class

Course : JEE

Start Date : 04.06.2023

End Date : 30.06.2023

Types of Batch : Live Online Class

 Subject M T W T F S S Physics(50 hours) 3:30 PM - - 3:30 PM - 3:30 PM - 