Question

0.5 m 30° A small o.05 kilogram block is released from rest at point A as shown above, a vertical distance h above the ground. It slides down an inclined track, around a circular loop of radius o.s meter, then up another incine that forms an angle of 309 with the horlzontal. The block slides of the track with a speed of 4.0 m/s at polnt C, which is a height of 0.50 meter above the Assume the entire track to be frictionless and air resistance to be negligible. a. Determine the heighth by t
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Answer #1

As the track is frictionless, the mechanical energy is conserved,

PE initial = PE final + KE final

PE initial = mgh

PE final = mgh1

KE final = 1/2 mv^2

mgh = mgh1 +1/2mv^2

h = h1 +v^2/2g
h = 0.5+16/(2*9.8) = 1.316 m

Free diagram for the block at the point B Block

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