Asked by Anonymous
A 62 kg person is skiing down a 37 degree slope that has a coefficient of 0.15. what is their velocity after 5.0 seconds
Answers
Answered by
Damon
weight = m g
normal force = m g cos 37
friction force up slope = .15 m g cos 37
proulsion force down slope = m g sin 37
net force
= m g (sin 37 - .15cos 37)
= m a
so
a = g(sin 37-.15 cos37)
calculate a from that
now
v = a t = 5 a
normal force = m g cos 37
friction force up slope = .15 m g cos 37
proulsion force down slope = m g sin 37
net force
= m g (sin 37 - .15cos 37)
= m a
so
a = g(sin 37-.15 cos37)
calculate a from that
now
v = a t = 5 a
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