This is part II of a three part problem (karma points for EACH of the three part

Question

This is part II of a three part problem (karma points for EACH of the three parts) Please see parts I and III also. Also, please don't just show the answer. Please also show very clearly how you found the answer.

A 350 kg coaster cart rests atop a 30m hill. When the cart is released it passes through a circular loop that has a 12m radius. Then, at constant velocity, the cart navigates a 10 m turn. (Assume no friction or air resistance.)
AFTER the car passes through (not during) the circular loop what will its velocity be?

Explanation / Answer

for the cart to avoid falling mg=mv2/r v2=rg=10*10 v=10m/sec needed now velocity of cart is given from .5mv2=mg*h,h=30 v=24.5m/sec so cart will not fall

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