Asked by Mike

A 52kg ice hockey player moving at 11m/s slows down and stops over a displacement os 8.0m.

Calculate the net force of the skater.
Answered by Steve
s = 11t + 1/2 at^2
8 = 11t + 1/2 at^2
at^2 + 22t - 16 = 0

v = 11 + at
at = -11
so,
-11t + 22t - 16 = 0
11t = 16
t = 16/11

a(16/11) = -11
a = -121/16

F = ma = 52(-121/16) = -1573/4 = -393.25N
Answered by Elena
Another method using the Work-Energy theorem
ΔKE =W(friction)
ΔKE = KE2-KE1 = 0 - mv²/2.
W(fr) = F•s,
- mv²/2 =F•s,
F =- mv²/2•s = 52•121/2•8 =393 N
Answered by Elena
F =- mv²/2•s = - 52•121/2•8 = - 393 N
Answered by gv
vgf
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