Question

A roller-coaster is at the top of a 62-meter hill. The car and its passengers have a total mass of 1,088 kilograms. By the time the car reaches the bottom of the hill, its speed is 33 meters per second. How much kinetic energy does the car have at the bottom of the hill?
A 592,416 J
B 2,978,944 J
C 17,952 J
D 1,184,832 J

Answer 1

To calculate the kinetic energy of the roller-coaster car at the bottom of the hill, we can use the formula for kinetic energy:

\[
KE = \frac{1}{2}mv^2
\]

where:
- \( KE \) is the kinetic energy
- \( m \) is the mass of the car and its passengers (in kilograms)
- \( v \) is the speed of the car (in meters per second)

Given that:
- \( m = 1,088 \) kg
- \( v = 33 \) m/s

Plugging in the values:

\[
KE = \frac{1}{2} \times 1,088 \, \text{kg} \times (33 \, \text{m/s})^2
\]

Calculating \( (33 \, \text{m/s})^2 \):

\[
33^2 = 1089
\]

Now substituting that back into the equation:

\[
KE = \frac{1}{2} \times 1,088 \, \text{kg} \times 1089
\]

Calculating \( \frac{1}{2} \times 1,088 \):

\[
\frac{1}{2} \times 1,088 = 544
\]

Now multiplying:

\[
KE = 544 \, \text{kg} \times 1089
\]

Calculating \( 544 \times 1089 \):

\[
KE = 592,416 \, \text{J}
\]

Thus, the kinetic energy at the bottom of the hill is:

\[
\boxed{592,416 \, \text{J}}
\]

So the correct answer is **A 592,416 J**.