Monica,
By any chance do you have a diagram of the problem? Unfortunately without a diagram we cannot tell where point A is.
A bead slides without friction around a loopthe-loop.
The bead is released from a height
of 18.9 m from the bottom of the looptheloop which has a radius 7 m.
The acceleration of gravity is 9.8 m/s^2.
What is its speed at point A ?
Answer in units of m/s.
2 answers
r=7 m
h=18.9 m
From the conservation of energy, we have
Ki+Ui=Kf+Uf
0+mgh=mv^2/2 +mg(2R)
v^2=2g(h-2R).
Therefore
v=sqrt(2g(h-2R))
=sqrt(2(9.8m/s^2)[18.9m-2(7m)]
=9.8 m/s
h=18.9 m
From the conservation of energy, we have
Ki+Ui=Kf+Uf
0+mgh=mv^2/2 +mg(2R)
v^2=2g(h-2R).
Therefore
v=sqrt(2g(h-2R))
=sqrt(2(9.8m/s^2)[18.9m-2(7m)]
=9.8 m/s
1 answer