A 55 kg roller coaster car is launched, from ground level, at 28 m/s. How fast will it be moving when it reaches the top of a loop, which is 12 m above the ground

The speed of the roller coaster when it reaches the top of the loop is 24 m/s

Explanation:

Hi there!

To solve this problem, let's use the conservation of energy and assume that there is no friction. In such a case, the kinetic energy is only converted into gravitational potential energy.

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the roller coaster.

v = speed.

The gravitational potential energy (PE) is calculated as follows:

PE = m · g · h

Where:

m = mass.

g = acceleration due to gravity.

h = height.

According to the conservation of energy, the mechanical energy of the roller coaster remains constant (no energy can be created nor destroyed). In this case, the mechanical energy (ME) is given by the sum of the kinetic and gravitational potential energy:

ME = KE + PE

Initially, PE = 0 because the roller coaster is on ground level (h = 0).

Then, initially, the mechanical energy is equal to the kinetic energy:

ME = KE = 1/2 · m · v² = 1/2 · 55 kg · (28 m/s) ² = 2.2 * 10⁴ J

When the roller coaster reaches the top of the loop at 12 m, some of the kinetic energy had to be used to acquire potential energy. Now, the mechanical energy can be expressed as follows:

ME = KE + PE

Now, the potential energy is not zero:

PE = m · g · h

PE = 55 kg · 9.8 m/s² · 12 m

PE = 6.5 * 10³ J

Since ME is constant, ME = 2.2 * 10⁴ J

Then:

ME - PE = KE

ME - PE = 1/2 · m · v²

2 (ME - PE) / m = v²

v = √ (2 (ME - PE) / m)

v = √ (2 (2.2 * 10⁴ J - 6.5 * 10³ J) / 55 kg)

v = 24 m/s

The speed of the roller coaster when it reaches the top of the loop is 24 m/s