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Consider a rollercoaster that begins from rest at a height (h) of 200 meters. At one point in the track, the coaster goes around a loop. At the top of the loop, the coaster is moving with a speed of 40 m/s. Calculate the radius of the loop. coaster A. 18.45 m B. 118.45 m C. 98.98 m D. 59.23 m

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Answer:

D. 59.23 m

Explanation:

Given that:

initial height  [tex]h_i[/tex] of the roller coaster = 200m

speed (v) = 40 m/s

use conservation of energy between starting height and top of loop

[tex]m*g*h_i = \frac{1}{2}mv^2 + m*g*h_f[/tex]

where

[tex]h_f[/tex] i.e the final height  = 2R

R = radius of the loop

Then

[tex]m*g*h_i = \frac{1}{2}mv^2 + m*g*2 R[/tex]

Divide both sides with m; we have :

[tex]g*h_i = \frac{1}{2}v^2 + g*2 R[/tex]

[tex]9.81*200 = \frac{1}{2}*40^2 + 9.81*2 R[/tex]

1962 = 800 + (9.81 )(2R)

1962 -800 = (9.81 )(2R)

1162 = 19.62 R

[tex]R = \frac{1162}{19.62}[/tex]

R = 59.23 m

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