A 67.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe", a track that is one-quarter of a circle with a radius of 2.55 m. What speed (in m/s) does he need at the bottom assuming that the local acceleration due to gravity is 9.80 m/s2?

7.41

To find the speed at the bottom of the quarter pipe, we can use the principle of conservation of mechanical energy. At the bottom of the track, the skateboarder only has kinetic energy, and at the top, the skateboarder only has potential energy.

The potential energy of the skateboarder at the top of the quarter pipe is equal to the kinetic energy at the bottom.

The potential energy at the top is given by:

PE = mgh

Where m is the mass of the skateboarder (67.0 kg), g is the acceleration due to gravity (9.80 m/s^2), and h is the height at the top, which is the radius of the quarter pipe (2.55 m).

PE = (67.0 kg)(9.80 m/s^2)(2.55 m)

Next, we can find the kinetic energy of the skateboarder at the bottom of the quarter pipe.

KE = 1/2 * mv^2

Where m is the mass of the skateboarder (67.0 kg) and v is the velocity at the bottom.

Since the potential energy at the top of the track equals the kinetic energy at the bottom, we can set the equations equal to each other:

(67.0 kg)(9.80 m/s^2)(2.55 m) = 1/2 * (67.0 kg) * v^2

Now, we can solve for v:

v^2 = (2 * (67.0 kg * 9.80 m/s^2 * 2.55 m)) / (67.0 kg)

v^2 = 2 * 9.80 m/s^2 * 2.55 m

v^2 = 50.196 m^2/s^2

v = √(50.196 m^2/s^2)

v ≈ 7.083 m/s

Therefore, the skateboarder needs a speed of approximately 7.083 m/s at the bottom of the quarter pipe to just make it to the upper edge.