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##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

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### Video Transcript

For this question, you have to use Newton's second law for that. Let me choose the following reference frame vertical axis that I'm calling. Why on a horizontal axis that I'm calling X now for the first item, we apply Newton's second law on the X axis that access you got that the net force is given by the mess off the skater times his acceleration next direction. The net force that is acting on these access is on Lee. That frictional force so minus because it's pointing to the negative side off the X axis. The frictional force Is it close to the mass of the skater times? He's acceleration, but the frictional force we know magnetic situation is given by they connect. Take no coefficients times the normal force. So this is minus the kinetic friction Local official times the normal force which is of course, to m times his acceleration with noticed that the vertical axis there are only two forces acting the weight and the normal and the skater using moving in the vertical axis, and it's not going to move. Therefore, the weight force on the normal force are equal forces, so the normal force in the situation is because of the weight force, then minus you kinetic times. The weight force is increasing the mass off the skater times, his acceleration, the weight forces given by the mass off the skater times, the acceleration of gravity. So miners ethic, frictional coefficient times, am times G is he goes to m time acceleration. We can simplify the masses to get an acceleration. All that's access that is given by minus the kinetic additional coefficient times acceleration of gravity, which you should remember is 9.8 meters per second squared approximately. Then it's his acceleration is minus 0.1 times 9.8, which results in minus 0.98 meters per second squared. The minus sign indicates that the acceleration is pointing to the left, so it's decelerating with a magnitude off 0.98 meters per second squared. Then, on the second item we have to complete, how far will he move before stopping completely? For that, we can use to reach Alice Equation, which tells us that the final velocity squared is equal to the initial velocity squared plus troop times, the acceleration times, the displacement after stopping completely his final velocity will be zero. His initial velocity is 7.6. Then we have. He is acceleration, which is negative. Tu minus trees points 0.19 times a displacement. Now he's served his equation or the displacement. So two times 0.98 times displacement with 7.6 squared, then the displacement is 7.6 squares, divided by true times 0.98 and these results in approximately 29.5 meters. So this is the distance he will travel before stopping completely.

Brazilian Center for Research in Physics
##### Andy C.

University of Michigan - Ann Arbor

##### Marshall S.

University of Washington

Lectures

Join Bootcamp