During the winter break you go sledding with your friends. The hill you are sled
ID: 1433372 • Letter: D
Question
During the winter break you go sledding with your friends. The hill you are sledding on is L = 12 m long (that's the distance along the slope, not the vertical height of the slope) and slopes at an angle of ? = 17 degrees. At the bottom of the hill is a long horizontal stretch of snow where you slow down and stop.
If you start with zero velocity at the top of the hill and you stop a distance d = 29 m from the bottom of the hill, what is the coefficient of friction, ?k of the sled on the snow? Assume it is the same on both the hill and along the horizontal.
Give your answer to at least three significant figures to avoid your answer being counted wrong due to rounding.
Explanation / Answer
Friction is times the normal force.
When you are on the hill, you can project your gravitational force into the rotated coordinate system with an x-axis parallel to the hill, and a y-axis perpendicular to it. The result is a vector:
(g sin 17°, g cos 17°).
Thus, along the hill, your driving force is mg sin 17° and your normal force is mg cos 17°. Your frictional force is just m g cos 17°, and it points opposite to the driving force of gravity.
Work is force times distance. Both of these forces perform work on you through the L1 = 12m distance of the hill, and this work (by the work-energy theorem) becomes your kinetic energy at the base of the hill, before you travel down the horizontal pathway. This work is just:
W1 = (m g sin 17° - m g cos 17°) * L1.
On the bottom of the hill, your normal force increases to be just mg -- the ground beneath you directly opposes gravity, rather than turning it into forward motion like the hill did. And there is no more driving force. If the coefficient of friction is still the same, then the work done by friction over this distance L2 = 29 m is just:
W2 = - m g * L2.
This must remove the kinetic energy that you just acquired, so that we can say:
(m g sin 17° - m g cos 17°) * L1 - m g * L2 = 0.
We can now divide through by m*g to simplify this:
(sin 17° - cos 17°) L1 - L2 = 0.
L1 sin 17° = ( L1 cos 17° + L2 )
= L1 sin 17° / ( L1 cos 17° + L2 )
= 0.086
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.