Asked by Melissa
A 50.0kg cyclist on a 10.0kg bicycle speeds up from 5.0m/s to 10.0m/s , the total kinetic energy before acceleratingis 750, and after acceleratingis 3000, how much work is done to increase the kinetic energy of the cyclist and the bicycle
Answers
Answered by
Elena
Work for changing KE of cyclist+bicycle is
W=Δ KE= KE2 –KE1=
=3000-750 = 2250 J,
for the cyclist:
W1=ΔKE1 = m1•v2^2/2 – m1•v1^2/2 =
=50•100/2-50•25/2 =1875 J,
for the bicycle:
W2=ΔKE2 = m2•v2^2/2 – m2•v1^2/2 =
=10•100/2 - 10•25/2 =375 J
W=Δ KE= KE2 –KE1=
=3000-750 = 2250 J,
for the cyclist:
W1=ΔKE1 = m1•v2^2/2 – m1•v1^2/2 =
=50•100/2-50•25/2 =1875 J,
for the bicycle:
W2=ΔKE2 = m2•v2^2/2 – m2•v1^2/2 =
=10•100/2 - 10•25/2 =375 J
Answered by
Anonymous
I am 4 years in the future
Answered by
anon
dude same
Answered by
Anonymous
me too
Answered by
Paul
We could also use:
Answered by
Anonymous
50.0 10.0
Answered by
Anonymous
I am 9 years in the future :) so cool
Answered by
.
i am 10 years in the future <3
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