A 120 kg roller coaster car has speed 10 m/s at point A, descends 10.0 m to point B, then ascends 6.0 m to point C (heights measured from B). Assuming no friction or air resistance, calculate the kinetic energy, gravitational potential energy, and total mechanical energy at A, B, and C, and state where the car is fastest.

Answer

Taking point B as the zero of gravitational potential energy and using $g=9.8\,\text{m/s}^2$, the total mechanical energy is constant at $E=17{,}760\,\text{J}$ at A, B, and C. At A: $KE=6{,}000\,\text{J}$ and $GPE=11{,}760\,\text{J}$; at B: $KE=17{,}760\,\text{J}$ and $GPE=0\,\text{J}$; at C: $KE=10{,}704\,\text{J}$ and $GPE=7{,}056\,\text{J}$. The car is fastest at B (the lowest point). The change in gravitational potential energy equals the negative of the change in kinetic energy: $\Delta GPE=-\Delta KE$.

Explanation

What you are being asked to track

You are tracking how energy changes as the car goes down and back up, assuming mechanical energy is conserved (no friction, no air resistance). That means gravitational potential energy converts to kinetic energy and back, but the total stays the same.

Choose a reference height for gravitational potential energy

It is simplest to set point B as the zero height.

Height at A relative to B: $h_A=10.0\,\text{m}$
Height at B: $h_B=0\,\text{m}$
Height at C relative to B: $h_C=6.0\,\text{m}$

Use $$GPE = mgh$$ with $m=120\,\text{kg}$ and $g=9.8\,\text{m/s}^2$.

Energy at point A

Kinetic energy: $$KE_A = \tfrac12 mv_A^2 = \tfrac12(120)(10^2)=6000\,\text{J}$$ Gravitational potential energy: $$GPE_A=(120)(9.8)(10.0)=11760\,\text{J}$$ Total mechanical energy: $$E_A=KE_A+GPE_A=6000+11760=17760\,\text{J}$$

Energy at point B (lowest point)

At B, $h_B=0$, so $$GPE_B = 0$$ Conservation of mechanical energy gives $$E_B=E_A=17760\,\text{J}$$ So $$KE_B = 17760\,\text{J}$$ (That is why the speed is greatest at B: kinetic energy is maximum when potential energy is minimum.)

Energy at point C (back up the track)

Gravitational potential energy at C: $$GPE_C=(120)(9.8)(6.0)=7056\,\text{J}$$ Total is still $17760\,\text{J}$, so $$KE_C = 17760-7056 = 10704\,\text{J}$$

Where is it moving fastest?

Fastest means largest speed, which means largest kinetic energy. Since $KE$ is largest at point B, the car is fastest at point B.

Showing “change in GPE equals change in KE” (with sign)

Between A and B: $$\Delta GPE = GPE_B-GPE_A = 0-11760 = -11760\,\text{J}$$ $$\Delta KE = KE_B-KE_A = 17760-6000 = +11760\,\text{J}$$ So $$\Delta GPE = -\Delta KE$$ The same relationship holds from B to C: $GPE$ increases by $7056\,\text{J}$ and $KE$ decreases by $7056\,\text{J}$.

Skills You Achive

conservation-of-energy kinetic-energy gravitational-potential-energy energy-transformation