Question

A roller-coaster car may be represented by a block of mass 89.0 kg. The car is released from rest at a height h = 60.0 m above the ground and slides along a frictionless track. The car encounters a loop of radius R = 10.5 m at ground level, as shown. How fast is the car traveling at the top of the loop? Round your answer to the first decimal place.

Answer 1

To determine the speed of the roller-coaster car at the top of the loop, we can apply the conservation of mechanical energy. The speed of the roller-coaster car at the top of the loop with radius R = 10.5 m and height h = 60.0 m above the ground is 14.3 m/s.

Step-by-step explanation:

To determine the speed of the roller-coaster car at the top of the loop, we can apply the conservation of mechanical energy. At the initial height (h = 60.0 m), the potential energy is converted into kinetic energy at the bottom of the loop and gravitational potential energy at the top of the loop. The total mechanical energy is conserved in the absence of non-conservative forces.

The potential energy at the initial height is given by
`\(PE_(initial)\)` = mgh, where m is the mass (89.0 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (60.0 m).

The kinetic energy at the bottom of the loop is
`\(KE_(bottom) = PE_(initial)\)`.

At the top of the loop, the kinetic energy is converted back to potential energy, so
`\(PE_(top) = KE_(bottom)\)`. The potential energy at the top is
`\(PE_(top) = mgh_(top)\)`, where
`\(h_(top)\)` is the height at the top of the loop (which is the radius of the loop).

Now, set
`\(PE_(top) = KE_(bottom)\)` and solve for the velocity at the top
`(\(v_(top)\))` using the kinetic energy formula KE =
`(1)/(2)mv^2\)`.

`mgh_(top)` =
`(1)/(2)mv_(top)^2`

Solve for
`\(v_(top)\)`:

`\[ v_(top) = \sqrt{2gh_(top)} \]`

Substitute the values for g and
`\(h_(top)\)` to find
`\(v_(top)\)`.

`\(v_(top)\)` =
`√(2 \cdot 9.8 \cdot 10.5)`

= 14.3 m/s

So, the speed of the roller-coaster car at the top of the loop is 14.3 m/s.