A 5.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t + (0.610 m/s3)t3. What is the magnitude of F when t = 4.00 s?

Video transcript

Welcome back everybody. We have a bucket that is at the bottom of the well. So let me go ahead and draw our bucket right here. And we are told that this bucket is suspended by a cable that doesn't really have a mask that we're concerned with. Now this cable is exerting a force on the bucket and we are going to No Tate that with f as a function of time and you'll see why in just a second. We are told that the mass of this bucket is 12 kg and we are also given the position of the bucket as a function of time. And we are asked to find the magnitude of the force at a time of five seconds. Well, using Newton's second law, we know that the force is equal to mass times acceleration. But since we are given position as a function of time, this is going to be our acceleration as a function of time, specifically at five seconds. But how do we find this acceleration? Well, we know that the derivative of position with respect to time is going to give us our velocity as a function of time and the derivative our velocity with respect to time is going to give us our acceleration with respect to time, which is exactly where we are going to plug five into. So let's go ahead and take those derivatives here Derivative of position with respect to time, which of course is equal to our velocity is going to be equal to. Well, the derivative of this first term right here, 1.2 T that's just going to be 1. times the derivative of the second term which is a little tough to see. Let me actually write it off, it's going to be 0.4 to one times T cube derivative of the second term is going to be three times 0.4 to one times t squared. So now let's find the derivative of velocity with respect to time, which is of course going to be our acceleration. This is going to be equal to well, the derivative of this term is just a constant. So that'll be zero plus this term right here, Once again using the power rule, we're going to have two times three times 0.4 to one T. Wonderful. Now that we have our acceleration, let's go ahead and plug in our value of five seconds for time, our acceleration evaluated at five seconds is equal to well two times three is six times 0.4 to one times our time of five seconds. When you plug this into your calculator, you get that acceleration is 12.63 m per second squared. Wonderful. Now that we have our acceleration and our mass, we are good to find the magnitude of the force at a time of five seconds. This is going to be equal to our mass times acceleration of five seconds which we have both those values are mass is 12 times our acceleration right here of 12.63 m per second squared, which when you plug into your calculator, you get 152 newtons or responding to our answer choice of a thank you guys so much for watching. Hope this video helped. We will see you all in the next one.