# Answer to Question #114448 in Mechanics | Relativity for Jason Grace

Question #114448
A roller coaster at a popular amusement park has a portion of the track that loops. Assuming that the roller coaster is frictionless, find the velocity at the top of the loop.

V1 = 0 m/s
height of the starting point (where V1 = 0) = 70m
height of the loop = 30m
V2 = ?
1
2020-05-08T16:06:06-0400

Condition: A roller coaster at a popular amusement park has a portion of the track that loops. Assuming that the roller coaster is frictionless, find the velocity at the top of the loop.

"V_1=0 (\\frac{m}{s})"

height of the starting point (where V1 = 0) "h_1=70 (m)"

height of the loop  "h_2=30(m)"

"V_2-?"

Solution: According to the law of conservation of energy:

"K_1+P_1= K_2+P_2", (1)

where "K_1" is a kinetic energy in the starting point, "K_1=0", because "V_1=0"; "P_1=mgh_1" is a potential energy in the starting point; "K_2=\\frac{mV^2}{2}" is a kinetic energy of the loop; "P_2=mgh_2" is a potential energy of the loop.

Rewrite (1) taking into account "K_1, K_2, P_1, P_2."

"mgh_1=\\frac{mV^2}{2}+mgh_2", (2)

where m will decrease. So:

"gh_1=\\frac {V_2^2}{2}+gh_2" (3)

Now, let’s find "V_2."

"\\frac {V_2^2}{2}=gh_1-gh_2"

"\\frac {V_2^2}{2}=g(h_1-h_2)"

"V_2^2=2g(h_1-h_2)".

So

"V_2=\\sqrt{2g(h_1-h_2)}" (4)

Substitute numbers from the condition and calculate "V_2":

"V_2=\\sqrt{2*9.8(70-30)}=\\sqrt{19.6*40}=\\sqrt{784}=28 (\\frac{m}{s})".

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