Bonnie and Clyde are sliding a 300 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 385 N of force while Bonnie pulls forward on a rope with 350 N of force. What is the safe's coefficient of kinetic friction on the bank floor?

Video transcript

Hi everyone in this practice problem, we're being asked to determine the coefficient of kinetic friction. We will have two porters transporting a 250 kg luggage box towards the train waiting at the platform, one of the porters pulls it with a rope from the front with a force of 280 Newton while the other pushes it from behind with a force of 245 Newton. We're being asked to determine the coefficient of kinetic friction experienced by the luggage box. If the box moves with a constant speed, the options given are a 0.42 B 0.12 C 0.24 and D 0.21. So we will use the particle model and consider the luggage box as a particle moving along the positive X direction. So I am going to start us off with creating or drawing a free body diagram of our luggage which will just be considered as a particle. So this dot right here represents our luggage and this two lines that I draw are the X and Y axis. So first, the luggage itself will experience the gravitational force FG going downwards and the normal force countering that gravitational force. Second, it will experience two different forces um pushing or moving the luggage forward, which will be F one which is Newton and F two, which is going to be 245 Newton countering those two, it will have a frictional force of K acting on the opposite direction. So all of this are going to be the uh forces acting upon our luggage. And the next piece of information that is important is that the box will move with a constant speed which will say that the F net will then be comes equals to zero. So in this case, we can apply Newton's first law as the box is moving with a constant speed. So Sigma FX will be equals to zero. So constant speed means the box is at equilibrium. So Sigma FX will equals to zero and the forces acting on the X directions are F one plus F two minus FK which in this case, F one is 280 Newton F two is 245 Newton and FK is FK and all of that will be equals to zero. So FK will essentially be equal to Newton after rearrangement. Next, we want to look into the net four Sigma Fy in the, the net force in the right direction. So Sigma Fy will be also equals to zero because the luggage is not going to be moving in the vertical direction in the Y direction, there will only be two forces which is N and FG and N minus FG will then equals to zero where the normal force N will be equal to M multiplied by G or the gravitational force FG. Calculating this, the mass is going to be 250 kg and the gravitational acceleration is just 9.81 m per second squared. And calculating this, the normal force will then come out to a value of 2.45 time stand to the power of three Newton. Next, we want to actually dive deeper into the frictional force. Now that we have obtained our normal force. FK will be equal to UK multiplied by normal force. And the UK, the coefficient of kinetic friction is what we are interested to find. So rearranging the UK UK will then equals to FK divided by N FK is uh 525 Newton, which we have calculated previously and the normal force is 2. times 10 to the power of three Newton. So the coefficient of kinetic friction will come up to a value of 0.214 rounding it down, we can obtain the kinetic friction coefficient to be equal to 0.21. So there we have the answer to this particular practice problem which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem with the coefficient of kinetic friction equals to 0.21. And that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one and thank you so much for watching. Bye bye.