In the figure, a block slides along a track from one level to a higher level, by

Question

In the figure, a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless until the block reaches the higher level. Over there, a frictional force stops the block in a distance d. The block's initial speed is 8.5 m/s; the height difference h is 1.00 m; and the coefficient of kinetic friction is 0.60.

In the figure, a block slides along a track from one level to a higher level, by moving through an intermediate valley. The track is frictionless until the block reaches the higher level. Over there, a frictional force stops the block in a distance d. The block's initial speed is 8.5 m/s; the height difference h is 1.00 m; and the coefficient of kinetic friction is 0.60.

Explanation / Answer

  Let:
m be the mass of the block,
v0 be its initial speed,
h be the height difference,
d be the stopping distance,
u be the coefficient of kinetic friction,
g be the acceleration due to gravity.

Equating the block's initial kinetic energy to its increase in potential energy and the work done against friction:
mv0^2 / 2 = mgh + umgd
v0^2 = 2g(h + ud)
h + ud = v0^2 / (2g)
d = [ v0^2 / (2g) - h ] / u
= [ 8.5^2 / (2 * 9.81) - 1.0 ] / 0.60
= 4.477.

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