Asked by Anonymous
a girl on a swing 2m above the ground at her highest point and 1m above the ground at her lowest point. what is the girls maximum speed?
Answers
Answered by
Steve
If I understand conservation of energy correctly, if her weight is mg, then the loss of PE going from 2m to 1m is mg Joules.
So, since that is converted to KE, we have
mg = 1/2 mv^2
2g = v^2
v = √19.6 m/s
So, since that is converted to KE, we have
mg = 1/2 mv^2
2g = v^2
v = √19.6 m/s
Answered by
Hassan Anis
Loss in P.E = Gain in K.E
mgh = 1/2mv^2
g(1) = 1/2v^2
2g = v^2
v = √19.6 = 4.427 m/s (Answered)
mgh = 1/2mv^2
g(1) = 1/2v^2
2g = v^2
v = √19.6 = 4.427 m/s (Answered)
Answered by
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