Okay, so in this problem we have a sled that is 25 kilograms being pulled by a rope with 80 Newton's and the friction coefficient mu is 0.050 and they're asking what's the force of friction. So as always we start with a force diagram and we know we've got gravity down here and that's 25 kilograms times our 9.8 Newton's per kilogram or 245 Newton's. We have no other vertical forces so we know that F normal is equal but opposite and we know we've got 80 Newton's going out this way and we've got some friction going back that way and we're not giving the amount of friction we're just giving mu so that we will have to solve for. So from our mu equals force of friction over force normal we can rewrite that as force of friction is mu F normal so 0.05 times 245 is 12.25 Newton's. We can write that in here as well now. Okay so we've got that the next question is what's the acceleration? So for acceleration you know it's F net over the mass and net force is going to be 80 minus 12.25 divided by our mass or 2.71 meters per second per second. And the last question says if it starts from rest how fast will it be going 1.5 seconds later so we also know v final equals a t plus v initial or v final equals 2.71 meters per second per second times time is 1.5 seconds plus 0 equals 4.065 meters per second because our seconds canceled out. So the last question we are asked what if we change this 80 Newton's at an angle so now we need to solve for our y component in our x component of that so sine of 25 equals y over 80 and cosine of 25 equals x over 80. Do that math hit pause and come back and we should get y is equal to 33.8 and x is equal to 72.5 the y component is going to change our normal force so in our new force diagram we still have 245 down but now going up we have the normal force is 245 minus our 33.8 and our pole here is now only 72.5 and now we see mu so we see our normal force is 211.2 multiplying that times mu gives us 10.56 there is our normal and our mu is still .05 10.56 new is our new force of friction and we are out of time.