Question
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
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
Anonymous
I am 4 years in the future
anon
dude same
Anonymous
me too
Paul
We could also use:
Anonymous
50.0 10.0
Anonymous
I am 9 years in the future :) so cool
.
i am 10 years in the future <3