A roller coaster starts at a point 30 m above the bottom of a dip with a speed o

Question

A roller coaster starts at a point 30 m above the bottom of a dip with a speed of 18 m/s. Neglect friction, the speed of the roller coaster at the top of the next slope, which is 15 m above the bottom of the dip will be....?

The answer is 24.7 to 25

What I have so far:

v1 = sqrt 2gh = sqrt 2*9.8*30 = 24.25

But, this doesn't take in the second slope, so:

v2^2 - v1^2 = -2gh

v2^2 = sqrt v1^2-2gh

v2^2 = sqrt 24.25^2 - (2*9.8*1h)

v2 = 17.15 m/s

What did I do wrong? SHould I have stopped at the first step, despite there being two slopes?

Explanation / Answer

By Conservation of energy

KEi+PEi =KEf+PEf

(1/2)mVi2+mghi=(1/2)mVf2+mghf

(1/2)*182+9.8*30 =(1/2)*Vf2+9.8*15

456=(1/2)Vf2+147

(1/2)Vf2=456-147 =309

Vf2=2*309 =618

Vf=24.86 m/s

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