Work Energy Principle – Kinetics of Particles Work and Energy – Engineering Mechanics hi friends we’ll discuss the kinetics of particles kinetics means analysis of motion considering masses of the particles and the forces which causes the motion this branch of dynamics is called as kinetics therefore whenever would study about kinetics we will deal with forces as well as masses of the particle in kinetics of particles there are three parts kinetics of particles one called as the D alembert’s principle kinetics of particles two work energy principle and kinetics of particles three where we’ll study impulse momentum collision between elastic bodies in this lecture we’ll study about work energy principle work energy Princip in short it is also written as WEP let’s assume when block is at rest on horizontal plane say this one is a block which is at rest on horizontal plane it will draw free body diagram of this block weight of this block acts in downward direction and we’ll get reaction which is a developer contact surface there will say this one is weight of this block which is say m into g here what will get normal reaction Rn acting on the block now if this block is subjected to the resultant force which adds in say horizontal direction say this one force which is acting on this block is acting in rightward direction say magnitude of force is say P or may be considered as F assuming frictionless surface resultant force acting in horizontal direction is in the rightward direction therefore as any force acts on a body a resultant force acts on a body body always accelerates as per Newton’s law as per Newton’s law F equal to mass into acceleration that means whenever any resultant force acts on a body say F on a body having mass equal to M kg body s alerts a meter per second square that means as on this block of mass M kg this force P is acting this block starts its motion in rightward direction and accelerates in rightward direction now as it accelerates in the rightward direction people represent here some position initially block is at this location now once force P is applied then it accelerates that means say one more position say position one again block moves in the rightward direction I will represent one more position a safe position 2 add body is subjected to acceleration and that is also in rightward direction velocity of block increases as it was in rightward direction therefore and this position one if i’ll represents a this one is a position of block which one is at a distance of say x1 from its initial position this is initial position now at position one I will represent position is say x1 and as it passes through position one it’s velocity will assume say v1 again force is continuously applied on the body therefore body is subjected to acceleration and body continues its motion in rightward direction as mass of this block is young as per Newton’s law F equal to Ma body continues or body oscillates in the rightward direction after some time next time instant body passes through this position to i’ll represent again here this position mass of block is M force is still acting on the body this one is for P now this location or this position I’ll represent here as position 2 which is at a distance of X 2 from its initial position and this block attends some velocity say V 2 as a position 2 it means initially force P is applied on the body as per Newton’s law block of mass M accelerates when it passes through position 1 it attains velocity V 1 when its position is X 1 from its initial location again motion continues due to its acceleration it attains some different velocity V 2 mass of block is M kg and its position from initial position is say X 2 now as per this equation F equal to M a infant this equation as F equal to mass into this acceleration equal rate in terms of velocity and in terms of displacement as V dv by dx aceleration is DV by DT multiplying and dividing throat by DX what will get DV by DX that means this equation is written as what F into DX equal to e M V DV where it represents what even for F acts on a body or here this is resultant force here what force is acting is P that means F equal to P for this small displacement DX velocity of block varies by this dv as force P is acting in rightward direction for small displacement DX change in velocity is dv now if you want to analyze from position 1 to position 2 – we’ll integrate this equation from position 1 to 2 where limits of you can say displacement which DX is varies from x1 to x2 and velocity varies from v1 to v2 therefore integrating this equation is the next page it will integrate integral of F DX equal to integral of M V DV it will integrate this one from x1 to x2 when position is x1 velocity is given and when position is of 2 that means x2 velocity is v2 therefore if you integrate now watergate EF into P X where X varies from X 1 to X 2 which is equal to integral of M V DV is M square by 2 that means half M v square where this velocity varies from v1 to v2 substituting your upper limit and lower limit what we’ll get here here upper limit is x2 minus lower limit is x1 equal to half M upper limit is V 2 V 2 square minus lower limit is V1 total related V 1 square F into this x2 minus x1 represents displacement of this block from position 1 to 2 therefore this one is also written as F into say Delta X which is equal to now this one becomes 1/2 M V 2 square minus 1/2 M V 1 square force into displacement is nothing but work done by this force when it it’s a displacement is Delta F therefore this one is written work done bay forces and displacement when body moves from position 1 to 2 therefore w1 to 2 equal to half MV square is kinetic energy what term you are getting is half MV 2 square therefore this one is kinetic energy of this block when it is at position 2 minus half MV 1 square represents kinetic energy at position 1 therefore this one is relation between work done by a full-size swing body moves from position 1 to 2 and that is equal to kinetic energy at position 2 minus kinetic energy at position 1 which represents a relation between work done and kinetic energy of this block having mass equal to M kg this relation is called as work energy principle normal type of problems that can be solved using work energy principle are what work return force into displacement kinetic energy half MV square it means if displacement is known we’ll find out maybe velocity at final position using this principle work done my process is k2 minus k1 what it represents when work done by forces when body moves from one to two he is final kinetic energy minus initial kinetic energy that means work done by processes change in kinetic energy all of these work then what is important this work done term is very important basically work done is always product of two terms it is force into displacement now this one general equation force into displacement but what is important here direction of force and displacement if force is horizontal and displacement is inclined then you have to consider displacement in the direction of force sometimes may be displacement horizontal force inclined so what force is acting in the direction of displacement therefore this one is written as what force into displacement in the direction of force force into displacement in the direction of force or this one is force in the direction of displacement force in the direction of displacement into displacement of law that means direction of force and displacement what we are getting here that must be same force vertical the displacement must be vertical when this work done is positive and when this work done is negative that is also important therefore what is important here if direction of force if direction of force and displacement direction of force and displacement are same if direction of force and displacement are same then work done is a positive then work done is positive sometimes may be you may get motion in some direction and force which is acting is in opposite direction which is always possible or always relegated for frictional force because it opposes the motion therefore you may get a negative worker also if direction of force and displacement if direction of force and displacement are opposite are opposite then work done then work done is negative that means while finding work done by any force or motion to concentrate force is acting in which direction and displacement of block is in which direction again we’ll discuss this part if you want to find out what I del for example say this one is a block which is resting initially on horizontal plane if we draw free body diagram weight will get in downward Direction say mg what will get here normal reaction now if block moves in rightward direction say this one is displacement but now what force is acting on the block is not horizontal if force acting on this block is say inclined force force P inclined at an angle of theta now displacement you know it is in horizontal direction but force P is inclined therefore now to find out what force is acting in the direction of displacement that means it will resolve this P what will get here P cos of theta therefore in this case work is done by this P cos of theta because it is acting in a horizontal direction into this displacement DX now other forces what we are getting here the component of P is acting in downward direction which one is equal to what P sine theta P sine theta in vertically downward direction mg means weight of block acting in downward direction normal reaction acting in upward direction what is the work done by P sine theta or mg or normal reaction all these three forces are acting in a vertical direction but displacement of block is in horizontal direction that means there is no displacement in verticle direction therefore work done by all these three forces P sine theta M G and Rn is 0 that means when we’ll get work done by any force as 0 if forces if forces acts perpendicular to the direction of motion perpendicular to the direction of motion then work done by forces equal to 0 let me always you know concentrate which force acts in the direction of displacement in the direction of motion the year only one force is acting in horizontal direction or in the direction of displacement and that is also direction of force and displacement say work done by only P cos theta and that work done is positive because direction of force and displacement are same similarly if I take here say a block which one moves along inclined plane inclined at an angle of theta say this one is a block having some weight W say this one is weight that means mg now if we try to find out work done by forces when this block moves in downward direction by this a displacement DX the motion is along the plane downward now it will represent various forces as inclination of plane is Theta with horizontal this angle what we give theta if will resolve you get one component here W cos theta one more component will read here which adds along the plane component is W sine theta normal reaction acts on this block say this one is what Rn if µ is given that means coefficient of friction is given you will get here frictional force which opposes the motion whose magnitude is equal to µ into Rn now work is done by which forces first direction of displacement is important displacement is along the plane downward the first you check which forces are acting along the plane along the plane you are adding two components one is W sine theta other one is µ Rn that means work done by W cos theta and Rn and R 0 you will get work done only by W sin theta and memory again displacement is in downward direction they are put work done by W sine theta becomes positive and work done by µ RN is negative because what you are µ frictional force is acting in opposite direction of motion in this case work done is equal to W sine theta into D X this one is positive and work is done by this frictional force is µ R n into this displacement is Dx now this µ R n means frictional force this always acts in opposite direction of motion therefore any problem whenever will gate this work done or any frictional force the work done by frictional force her done by frictional force is negative why it is always negative because direction of force frictional is always oppose it to the direction of motion thank you

RELATED ARTICLES     kanishk bansal says:

sir , how we will say that the friction force is acting on a body. Aksh Patel says:

Background me kuch avaz aa Raha he so not audible properly Sai Charan says:

Jst loved it..great explaination..but if the noise is reduced..then it would be even better..😀 Nikhil Nirala says:

Pronounciation of 'M' is yum😕 very irritating 😤 lilnimo says:

Yef RahulByte says:

Watch this at speed 1.25.. Thank me later Kuchay Mehak says:

Best lectures ……hats off kartik thakur says:

Yum yuf Ekeeda says:

Hello Friends,

Watch Complete Video Series of Subject Engineering Mechanics only on Ekeeda Application.

Use Coupon Code "NEWUSER" and access any one Subject free for 3 days! ankita kashyap says: