 Hello everyone, myself, Sandeep Javeri, Assistant Professor, Department of Civil Engineering from Valjean Institute of Technology, Sholapur. Now in today's session, we are going to discuss a problem on principle of conservation of energy for the pile hammer problem. Now the learning outcome, at the end of this session students will be able to solve problem on law of conservation of energy. Now before solving the problem we must understood what is the law of conservation of energy. According to that the energy can neither be created nor destroyed, though it can be transformed from one form to another form. That means the total energy which is possessed by a body remains constant provided no energy is added or taken from it. For example, if you consider the water which is stored in a tank, say it is a RCC tank, so the height of the tank is H. Now if you open this knob or water tap, then the water start flowing down, so this is by gravity. That means water the water is stored in a tank before opening the tap, this energy is stored that means it is a potential energy and when you open the tap the water start flowing that means the water which is as stationary now it is in moving position that means the energy is converted from potential energy to kinetic energy. So this is the best example of showing the law of conservation of energy. Now let us took a problem on pile hammer, a hammer of mass 400 kg falls through height of 3 meter on a pile of negligible mass, if it drives the pile 1 meter into the ground find the average resistance of the ground for penetration. So to solve this problem we have to consider how the system is works. So let us consider this figure, here pile is shown having mass is small m, now the mass of hammer is capital M, the hammer having initial velocity is 0, the height of the hammer that is pile hammer from the pile is given as 3 meter. Now here we have to consider the weight of hammer that is 400 kg, the pile having negligible mass, so we are not considering the mass of the pile, the height of the hammer from the pile is 3 meter and it drives 1 meter into the ground and obviously the ground will have some resistance for the penetration of this pile, so that resistance we are going to find out, the penetration distance that is they have given 1 meter. Now let us consider m is a mass of the hammer as 400 kg, height through which hammer force is h that is equals to 3 meter as shown in the figure. Now let us consider the velocity of hammer before striking the pile is given by a small v meter per second and depth of penetration of hammer that is equals to 1 meter that is given by the letter s, mass of the pile is negligible so small m value is work out to be 0. Now step number one using the expression for freely falling body, as we know for freely falling body we can get the velocity by using the expression v square minus u square is equal to 2 gh, if you consider initial velocity u as 0 then we are getting the expression v is equal to under root 2 gh or you can say 2 gh raised to half, where h is the height of the pile hammer from the pile g is accession due to gravity v is the final velocity of the hammer. So by putting the value of g as 9.81 and the h as 3 meter we are getting the final velocity of the hammer as 7.672 meter per second. Step number two after striking the hammer on the pile what will happen this pile and this hammer they are moving with the common velocity. So let us consider that common velocity is capital V. Now how to get that common velocity that is a question. Now we know there is a law of conservation of momentum that the initial momentum before impact is equal to final momentum after impact. So we are going to use that concept and writing the expression for this. So m into small m into 0 that 0 is nothing but initial velocity of the pile which is 0 obviously it is at rest then plus the initial momentum of this pile hammer which how the velocity now is small v that is 7.672 meter per second the mass of the pile hammer is given as 400. So 400 into 7.672 that is equals to addition of combined masses of pile hammer and pile. As we know the weight of pile is negligible that is 0. So we are not considering that weight. So mass of the hammer is 400 and it is multiplied with the common velocity v. So this is the expression for initial momentum equals to final momentum that is nothing but obeying the law of conservation of momentum. So we are getting the common velocity after impact that is 7.672 meter per second. Now step number 3 the kinetic energy of the system how it is to be work out. So for that we are considering the expression 1 half capital M plus small m v square that is equals to half mv square which is equals to half into 400 into 7.672 square. So m is a mass of the pile hammer as the mass of the pile is negligible. So we are considering here mass of the pile hammer. So this kinetic energy of the system will work out to be 11771.9 Newton meter. The potential energy of the system lost in moving a depth of one meter that is equals to we have expression mgh the expression capital M plus small m into g into s. So putting the value of s that is a depth of penetration as one meter and g as 9.81 the mass of the hammer is 400. So we are getting the potential energy as 3924 Newton meter. The total energy of the system it is a combination of or addition of kinetic energy plus potential energy that is equals to 11771.9 plus 3924 that is equals to 15695.9 Newton meter. Now the work is done during the penetration of the pile. So if you consider the figure here this is a penetration of this pile having mass is m which is negligible. So s is given as 1 meter and there is a resistance offered from the ground that is r and our target is to find out what is the resistance offered by this ground. For that we have considered the total energy concept and we know we have worked on during penetration of the pile is resistance of the ground multiplied by depth of the penetration. So this is the resistance and this is the depth of the penetration. We have r is unknown s is displacement that is 1 meter it is also a penetration or displacement of a pile or penetration of the pile into the ground. Now from principle of conservation of energy we have worked on is equal to total energy lost. So as we know r is unknown multiplied by 1 that is a distance or penetration then equals to total energy lost that is 15695.9 that is coming from the equation 1 and 2. So by solving this we are getting the value of resistance to penetration as 15.659 kilo Newton. So this is the way how to get the resistance of the ground for penetration for the given pile hammer problem. Now you are supposed to pause this video and answer this question. The question is hammer of mass 750 kg drops height of 1.2 meter on a pile of mass 200 kg. Find the velocity with which the hammer strikes on the pile. So in this problem they have given mass of the pile also and mass of the hammer also the height from which it falls on the pile that is also given as 1.2 meter. So these are the options given. Let us see how to solve this. This is a cross section of the pile hammer having necessary velocity is 0. The mass of the pile hammer is 750 kg. This mass of the pile is 200 kg. As we know for freely falling body the final velocity is given as 1002 gh or v is equal to 2 gh raise to half that is equals to 2 into 9.81 into 1.2 raise to half. So h is 1.2 meter and g is 9.81 meter per second square. So velocity that is final velocity will be 4.85 meter per second. So the right option is 4.85 meter per second. Now these are the references which we are using for creation of this video. Thank you.