JG-3 A 2.50kg block is sliding at 5.75m/s down a rough surface for which -0.350 and μ-0.700. The surface is inclined 15.0° from the horizontal. The block eventually comes to a stop after sliding a distance S down the incline. For this problem neglect the effect of air resistance. Find the distance S, in meters. Mechanical energy will not be conserved, so use WNC = AK+AU. V=5.75m/s S stopped 15.0°

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Chapter1: Units, Trigonometry. And Vectors
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AUG-3 A 2.50kg block is sliding at 5.75m/s down a rough
surface for which -0.350 and μ-0.700. The
surface is inclined 15.0° from the horizontal.
The block eventually comes to a stop after sliding
a distance S down the incline. For this problem
neglect the effect of air resistance.
Find the distance S, in meters. Mechanical energy
will not be conserved, so use WNC = AK + AU.
V=5.75m/s
S
stopped
15.0°
Transcribed Image Text:AUG-3 A 2.50kg block is sliding at 5.75m/s down a rough surface for which -0.350 and μ-0.700. The surface is inclined 15.0° from the horizontal. The block eventually comes to a stop after sliding a distance S down the incline. For this problem neglect the effect of air resistance. Find the distance S, in meters. Mechanical energy will not be conserved, so use WNC = AK + AU. V=5.75m/s S stopped 15.0°
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