Question

A 600000 g rollercoaster car starts at rest atop a 38 m hill Find the ( a ) gravitational potential energy ( b ) kinetic energy , ( c ) mechanical energy , and ( d ) speed of the car when it goes over the second hill which is 14 m high . You may assume the track is frictionless and that air resistance is negligible .

Answer 1

Given:

Mass, m = 600000 g

Height, h1 = 38 m

Height, h2 = 14 m

Let's solve for the following:

• (a). Gravitational potential energy

To find the gravitational potential energy, apply the formula:

`_GPE=m*g*h`

Where:

m is the mass in kg = 600000 g = 600 kg

g is the acceleration due to gravity = 9.8 ms/s²

h2 is the height in meters over the second hill = 14 m

Input the values into the formula and solve for PE:

`\begin{gathered} _GPE=600*9.8*14 \\ \\ =82320\text{ J} \end{gathered}`

Therefore, the gravitational potential energy at the second hill is 82320 Joules

• (b),. Kinetic energy at the second hill

To find the kinetic energy, apply the formula:

`KE=(1)/(2)mv^2=mg\Delta h=mg(h_1-h_2)`

Where:

m is the mass

h1 = 38 m

h2 = 14 m

Thus, we have:

`\begin{gathered} KE=600*9.8(38-14) \\ \\ KE=600*9.8(24) \\ \\ KE=141120\text{ J} \end{gathered}`

• (c). Mechanical Energy.

To find the mechanical energy, apply the formula:

Mechanical Energy = Potential energy + Kinetic energy

Mechanical Energy = 82320 + 141120 = 223440 J

• (d). Speed of the car when it goes over the second hill which is 14 m.

To find the speed of the car, apply the formula:

`KE=(1)/(2)mv^2`

Where:

KE is the kinetic energy = 223440 J

m = 600 kg

v is the velocity.

Let's solve for v:

`\begin{gathered} 223440=(1)/(2)*600*v^2 \\ \\ 223440=300v^2 \\ \\ v=\sqrt{(223440)/(300)} \\ \\ v=27.29\text{ m/s} \end{gathered}`

The speed when it goes over the second hill is 27.29 m/s

ANSWER:

• (A). 82320 J

,

• (b). 141120 J

,

• (C). 223440 J

,

• (d). 27.29 m/s