David P. answered 05/04/15
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PhD with teaching and tutoring experience, Math/Science
This question deals with conservation of energy. Because the friction is assumed to be zero, we only have to consider potential energy and kinetic energy (no work is being done by or on the system). As the roller coaster car goes up the hill its potential energy increases and its kinetic energy decreases, but the total amount of energy remains the same. So the change in potential energy of the car going from the bottom of the hill to the top plus the change in kinetic energy will be zero.
ΔKE + ΔPE = 0
The equation for kinetic energy is (1/2)*m*v2
The equation for potential energy is m*g*h
So, we get:
ΔKE = KET - KEB = (1/2)*m*(vT)2 - (1/2)*m*(vB)2
ΔPE = PET - PEB = m*g*hT - m*g*hB
The subscripts T and B correspond to values at either the Top or Bottom of the hill.
Let's set the height of the bottom to be our zero point (for simplicity). Thus our change in potential energy is equal to m*g*hT and our equation becomes
ΔKE + ΔPE = 0
(1/2)*m*(vT)2 - (1/2)*m*(vB)2 + m*g*hT = 0
Notice that the mass, m, is in front of every term. So we can divide both sides by m and eliminate it from the equation.
(1/2)*(vT)2 - (1/2)*(vB)2 + g*hT = 0
Now divide by g:
(1/2)*((vT)2 - (vB)2)/g + hT = 0
And finally move the velocity terms to the other side of the equation.
hT = -(1/2)*((vT)2 - (vB)2)/g
Insert the values and solve for the height at the top of the hill.
hT = -(0.5)*( (3 m/s)2 - (8.2 m/s)2)/(9.8 m/s2)
hT = 2.97 m
NOTE: The question is a bit poorly worded. Since the coaster must have a speed of at least 3 m/s at the top of the hill, it should ask what is the maximum height the hill can have in order for the roller coaster to work properly.