Asked by jake carrey
What constant horizontal force, acting over 16 m of level trail, gives a friction-free 51-kg skier, starting from rest, a speed of 5.7 m/s?
Answers
Answered by
Steve
v = at
5.7 = at
a = 5.7/t
s = 1/2 at^2
16 = 1/2 (5.7/t)t^2
32/5.7 = t
t = 5.61
a = 1.02
F = 51*1.02 = 52
5.7 = at
a = 5.7/t
s = 1/2 at^2
16 = 1/2 (5.7/t)t^2
32/5.7 = t
t = 5.61
a = 1.02
F = 51*1.02 = 52
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