Question

1. (varsity physics - a hard one! see 8 and 9 on page 1) calculate the following:

a. calculate the potential energy of a skater when they are at a starting height of 6 m.
b. if all of the potential energy the skater has at the top of the ramp is converted to kinetic energy at the bottom of the ramp (0 meters high), what would be the skater’s velocity at the bottom of the ramp? (hint: use the equations for gravitational potential energy and kinetic energy and set them equal to each other, then solve for velocity).
after you have calculated the velocity, check your work by trying this with the setup above, or by going to the intro simulation (friction off, grid on, speed on, slow setting) with a track shaped like this: \\ start the skater (red dot) at a height of 6 m and observe the speed when they roll off the ramp.

1. explore other options in the playground simulation. (choose one)

a. what happens if you turn off \stick to track\? what happens to the kinetic, potential, and thermal energy with that turned off?
b. try making a loop in the track. does your skateboarder make it around the loop? what if you turn on friction? what factors determine whether the skater makes it through the loop?

Explanation:

Part 3a

Step1: Recall Potential Energy Formula

The formula for gravitational potential energy is \( PE = mgh \), where \( m \) is mass, \( g = 9.8\ m/s^2 \) (acceleration due to gravity), and \( h \) is height. Since the mass of the skater is not given, we assume a typical mass (e.g., \( m = 60\ kg \), a common skater mass) for calculation.

Step2: Substitute Values

Substitute \( m = 60\ kg \), \( g = 9.8\ m/s^2 \), and \( h = 6\ m \) into the formula:
\( PE = 60\ kg \times 9.8\ m/s^2 \times 6\ m \)
\( PE = 60 \times 9.8 \times 6 \)
\( PE = 3528\ J \) (If mass is unknown, the answer can be left in terms of \( m \): \( PE = 58.8m\ J \))

Step1: Set PE = KE

Gravitational potential energy (\( PE = mgh \)) converts to kinetic energy (\( KE = \frac{1}{2}mv^2 \)). So, \( mgh=\frac{1}{2}mv^2 \).

Step2: Solve for Velocity

Cancel \( m \) from both sides: \( gh=\frac{1}{2}v^2 \).
Multiply both sides by 2: \( 2gh = v^2 \).
Take square root: \( v=\sqrt{2gh} \).
Substitute \( g = 9.8\ m/s^2 \), \( h = 6\ m \):
\( v=\sqrt{2\times9.8\times6} \)
\( v=\sqrt{117.6} \)
\( v\approx10.84\ m/s \)

When “stick to track” is off, the skater leaves the track (e.g., at the top of a loop or steep slope). Kinetic energy (\( KE \)) depends on speed, potential energy (\( PE \)) on height, and thermal energy (from friction) is minimal (friction is off in the setup). Without “stick to track,” the skater’s path is projectile - like: at launch, \( KE \) (from speed) and \( PE \) (from height above ground) change as they move, but mechanical energy ( \( KE + PE \)) is conserved (no friction). Thermal energy remains near zero (friction off).

Answer:

If \( m = 60\ kg \), potential energy is \( 3528\ J \) (or \( PE = 58.8m\ J \) if mass is variable).

Part 3b