 Hello friends, in today's session, I am going to deliver lecture on work energy principle problem. So, we are going to solve one problem on work energy principle, myself, Bipin Patti, assistant professor, department of civil engineering, Vulture, Institute of Technology, so on. So, these are the learning outcomes. So, at the end of this particular session, learner, students will be able to solve the problem on work energy principle. So, directly, we will discuss one problem, a block having a weight 2500 Newton, so simultaneously you just observe this particular figure, when block is present, and the weight, the mention in your problem statement, 2500 Newton, rest on a horizontal level plane, so this is nothing but your horizontal level plane, for which coefficient of friction means in between these two particular, so this is your plane phase of this particular horizontal block. And the in portion of this particular ground surface or whatever the material is there, so the coefficient of friction between these two particular materials, they are mentioned here 0.2. This block is pulled by a force of 1000 Newton, so this particular force they are mentioned here 1000 Newton, acting at an angle of 30 degree, so this particular angle they are mentioned with respect to horizontal, find the velocity of a block after it moves 30 meters starting from a rest, so the displacement dimension here 30 meter, and it starts from a rest, first case is the second case, if the force of 1000 Newton is then removed, means this additional force if we remove, then how much further will it move means you have to identify the displacement, so this force and additional force they are mentioned here. Now our first duty is try to prepare free body diagram for this particular figure, so this is the free body diagram of a block, show the reactions, in vertical downward direction weight is present, so opposite direction resultant, that is nothing but the reaction force is present, here the additional load 1000 Newton they are mentioned it is an incline, so show the component x and y component, so x component 1000 cos 30, y component 1000 sin 30. Now this particular block it moves in x direction means in horizontal direction, so the frictional force is developed exactly in opposite direction, this is called as free body diagram of a given block, now first step consider work energy of a given block and apply this particular principle and try to identify the unknown values, so first one summation v is equal to 0, all vertical forces were equating to 0, so r vertical upward force positive minus vertical downward force 2005 Newton plus 1000 sin 30 is equal to 0, now here r is unknown calculate the value of r 2000 Newton, by considering this particular figure and this equation we identify the value of r, second use friction law, by considering friction law f is equal to mu into r, so frictional force is just calculated r the reactions we already calculated mu coefficient of friction in your problem statement they are given that is f is equal to 0.2 into 2000, so that 2000 is nothing but this particular force, so the value and the answer regarding this particular frictional force is 400 Newton give the equation number 2. Now third step apply summation h is equal to 0 means energy principle is just apply all horizontal forces is use a security to 0, so by using energy principle r into s is equal to m by 2 into v square minus u square, so work done is equal to change in kinetic energy is there, so r is nothing but you just consider all horizontal forces, so in your problem statement s they are mentioned 30 meter m mass is there and they are mentioned the force, so we just convert this particular force into mass, so 2500 Newton divided by 9.81, so that is nothing but your mass and this mass is just put here that is nothing but the calculation, so r right hand side force is positive 1000 cos 30 minus frictional force I have already calculated 400 Newton minus 400, so that is nothing but your r into 30 that is s is there already they are mentioned in your problem statement this force is just convert into mass, so 254.84 means 2500 divided by 9.81 divided by 2 into v square minus 0 because initial velocity is 0, so the final velocity you just calculated and the value it comes 10.47 meter per second. So, in this one the condition they are mentioning in your problem statement consider work energy of a block after removing 1000 Newton force, so whatever the inclined force is present here in your problem statement remove it and try to identify the remaining value that is the displacement, so see here from friction law f is equal to mu into r has a similar formula is there, f is equal to 0.2 into r is here by considering this particular block, so the value of r is 2500 Newton f is equal to 500 Newton, now we just apply the principle to this particular block r into h is equal to m by 2 v square minus u square now the initial velocity is 0 in previous case, now here final velocity is 0 and the initial velocity you just consider according to this particular equation first case, so r is equal to left hand side force is negative minus f into s is equal to m by 2 v square minus u square, so f you already calculated minus 500 into s that is unknown m force you just convert into mass 254.84 divided by 2 final velocity is 0 initial velocity already were calculated this value put here. Do the calculation after the calculation just identify the displacement s is equal to 27.94 meter, so according to that easily we can calculate by applying this energy principle easily we can calculate initial velocity, final velocity, the displacement and remaining unknown values. Now my dear friends you just try to pause this video read this particular question and try to give the answer of this particular equation which is the following factor related by work energy principle, so this is the answer of this particular question force, velocity and displacement, displacement s, velocity initial final and force that is very important, so this is the correct answer. To prepare this particular session I refer this particular references, thank you.