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
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.
4 answers
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
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
F =- mv²/2•s = - 52•121/2•8 = - 393 N
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