Hello everyone and welcome to Gem Academy. In this example we have a block of mass m being pooled to the right by a force that is given by alpha or multiply by t. In other words the force is a function of time. We are asked to calculate how far will the box travel after 5 seconds. Let's start by doing our free body diagram. We have the weight at vertically downwards. We have the force F and definitely we have kinetic friction F key acting to the left. On one last force we have the normal force acting vertically upwards from the bottom of the block. The next step is to ask to apply Newton's second law equations in both directions. I am going to start with the y direction. The summation of Fy is equal to N minus mg. This is equal to 0. This means that N is equal to mg. This is the value of the expression for the normal force acting on the box. Now the summation of F of X is equal to F minus Fk. This will be equal to ME. But what do we know? By definition Fk is equal to mu k which is just mu multiplied by M by N. This means that the frictional force on the box is given by mu mg. Therefore we can conclude that alpha t minus mu mg is equal to ME. Alpha t minus mu mg is equal to ME. So how do we proceed from here? Remember that A is equal to Dv over Dt. If A is equal to Dv over Dt this means that Mdv over Dt is equal to alpha t minus mu mg. We can choose at this particular moment to put in the values M is equal to 4. So this is 4dv Dt equal to 2 because alpha is a constant 2. This is minus 1 half multiplied by 4 multiplied by G which is I'm going to take G to be 10 even though it's 9.8. So we will have 4dv over Dt equal to 2t minus that will be 4 divided by 2 is 2 minus 20. Hence we have Dv Dt equal to 0.5t minus 5. Or you can choose to use half minus 5. What that means is that Dv is equal to half t minus 5 multiplied by Dt. If we take the integrals of both sides we are integrating from, remember that the initial velocity of the block is 0 from 0 to V. We are integrating from 0 to T. What that will give us is this would mean that V is equal to 1 over 4t squared minus 5t we are integrating from 0 to T. In other words this would mean that V is equal to 1 ford T squared minus 5t. This is the expression for the velocity of the box to the right as a function of time. Now this would imply that the x over Dt is equal to 1 ford T squared minus 5t which means that the x is equal to 1 over 4t squared minus 5t all multiplied by Dt. If we take the integrals of both sides we are integrating from 0 to x from 0 to T. We will have here x equal to 1 over 8t squared minus 5t squared. This is the expression for the position of the box as a function of time. The next question is just to substitute the time 5 for the time. If we do that x will become 1 over 12 minus 5 over 2. Now you can use your calculators to calculate this expression. You will be required to insert the answer in the question right below this video in our online management system. If you have any question on this please do not hesitate to ask and I will be able to help you. Thank you so much for your time and remember this. Bye bye.