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lapo4ka [179]
3 years ago
13

A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert

ed by the seats on the passengers is equal to ¼ of their weight.
Physics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

v=10m/sec

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

F_N+Fg

where

F_N =forceof the normal

Fg= force due to gravity

Generally the net force is given to be FC(force towards center)

F_C =F_N + Fg

F_N =F_C -Fg

F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

m_g + \frac{8m}{4}= \frac{mv^2}{8}

\frac{5mg}{4} =\frac{mv^2}{8}

v^2 =\frac{40*10}{4}

v=10m/sec

Therefore the speed of the roller coaster is given ton be v=10m/sec

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<h3>What is frictional force?</h3>

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brainly.com/question/1714663

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Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

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