In this problem on work and energy were told that a sled with a given mass starts up an incline with a given speed, were also given the coefficient of friction and asked to calculate. Firstly, how far up the incline this little travel, the condition that must be placed on the coefficient of static freshen static friction so that this lead does not get stuck at its highest point as well as the speed. That thesis lead will return back to its starting point. So firstly, we'll have a look at our diagram. So we have a frictional force acting down the slope, the normal force acting, acting perpendicular to the slope as well as the weight acting vertically downwards. Now, from this free body diagram, we can see that these some off the forces acting on the sled in our wide election is equal to be normal force f n minus the component off the weight perpendicular to the plane which is M g Constanta, and that's equal to zero. And so from here, we can see therefore, that the normal force F n balances the component off the weight perpendicular to the plane, so FN is good energy cost data. There is no motion in our Y direction and therefore the frictional force, if if are, is equal to be quite efficient off kinetic friction UK times, mg times the co sign off data. So that's the frictional force acting parallel to the plane so we can apply the conservation of energy with a frictional force. And we'll use subscript one to refer to the state of the start of the motion. And subsequent to refers to this lead at the top of its motion. So if we take the starting position to be the zero for gravitational potential energy, we have V one is equal to 2.4 m the second why one is equal to zero and the final velocity at the top of its motion V two is equal to zero. So we know that why, to its final position is equal to d sign Peter. And we need to solve for D where D is the distance it travels at the slope. So, using the conservation of energy, we have Yeah. Mm hmm. Conservation of work and energy. The initial energy e one is equal to he final mechanical energy. Last he walked done against friction, which is the frictional force times The distance has led travels deep, and so the simple eyes are half and the one squared plus M g Y one is equal to the final kinetic energy a half and the two squared plus M. G. Why to must you walk down against friction? Yes, friction Times D. So if we rearrange this equation and remember that the initial position, why one is zero on the final velocity V 20 we can solve for the distance up the slope D and D is the one squared, divided by to G into Yeah, sign Peter plus coefficient of kind of friction. UK Times Co. Sign of data and these values are all known so we can calculate D. So this is the initial velocity. 2.3 m per second. That's squid divided by to times gravitational acceleration. 9.8 m square. Second, Tang's sign off 28 degrees. That's 0.25 call sign 28 degrees. And so if we calculate this, we get the distance traveled up the slope to be 0.3910 meters, and if we keep our significant figures. This is 0.39 meters. So that's all. First answer for part A. Now for. Pardon me. We want to calculate where to find the condition that must be placed on the coefficient of static friction to prevent the sled from getting stuck at its highest point. Now for this led to to slide back down. The fiction Force will now point up the hill in the free body diagram. In order for the sled to slide down, the component of gravity along the hill must be larger than the maximum force off static friction. So this means M g flying Tita once be greater, then the frictional force, which means M g scientists must be created them U S times MGI can the coast side of teacher. And so rearranging this we see that the coefficient of static friction new s must be less than 10 off 28 degrees and therefore the coefficient of static friction as the restraint that it has to be less than 0.5 three and then we place it a strange on the coefficient of static friction so that the sled we'll slide back down after reaching its maximum distance up the slope. Now finally, in part C. If the sled slides back down, we want to know its speed when it returns to its starting point. So we can again apply the conservation of energy and include the work done by fiction. But subsequent one this time refers to the at the top of the incline and subsequent to would refer to this right at the bottom of the incline. So we know that in this case, V one is equal to zero. Why one is equal to D sign pita. Yeah, and why to well, obviously equal to zero at the bottom of the hill. So again, applying the conservation off energy we have at e one is equal to you too. Plus the work done by fiction, which is the fictional force times the distance deep. So the this can be written as a half m v ones where the initial kinetic energy last day initial potential energy which is m g y one, must equal to the final Connecticut G a half and the two squared plus m g. Why to last you were planning instruction? Yeah, if Times D now we want to solve of the two in this problem. So if we rearrange and substitute our values, we get V two. Remember, the one is equal to zero as well as y. Tu V two is equal to two times the square root of the square would rather off two times g d into sine theta minus new k cosign Tita. And so now we can put on that is into this and this is he square root off to times 9.8 m per square. Second, it's the best units here. Times D, which is 0.3910 times this sign of 28 degrees minus 0.25 Call sign 28 degrees and so we get the final velocity when the sled reaches its position at the bottom of the hill to be 1.4 meters per second.