Asked by Andrea
A 50kg skier starts from rest from the top of a 100m slope. What is the speed of the skier on reaching the bottom of the slope? Neglect friction.
would i do this:
=PE + KE
=mgh +1/2mv^2
=50(10)(100) + 1/2 (50)v^2
=50000 +25v^2
=2000 =v^2
44.7m/s =v
Is this correct? Thank in advance for you help
would i do this:
=PE + KE
=mgh +1/2mv^2
=50(10)(100) + 1/2 (50)v^2
=50000 +25v^2
=2000 =v^2
44.7m/s =v
Is this correct? Thank in advance for you help
Answers
Answered by
bobpursley
I would do it like this, I do not understand your algebra.
FinalPE+ FinalKE= InitialPE+ initialKE
0+ 1/2 50 v^2= 50*9.8*100 + 0
v^2= 2*50*9.8*100
which is about what you got. However I used 9.8m/s^2 for g, which is much closer to what it actually is. Nowhere on Earth is it 10m/s^2, nor close to it.
FinalPE+ FinalKE= InitialPE+ initialKE
0+ 1/2 50 v^2= 50*9.8*100 + 0
v^2= 2*50*9.8*100
which is about what you got. However I used 9.8m/s^2 for g, which is much closer to what it actually is. Nowhere on Earth is it 10m/s^2, nor close to it.
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