In the physics laboratory, a glider is released from rest on a frictionless air track inclined at an angle. If the glider has gained a speed of 23.3 cm/s in traveling 60.5 cm from the starting point, what was the angle of inclination of the track?

To solve this problem, we can use the principle of conservation of energy. The potential energy at the start is converted into kinetic energy at the end.

1. First, we need to convert the given values to SI units. The speed v = 23.3 cm/s = 0.233 m/s and the distance d = 60.5 cm = 0.605 m.

2. The initial potential energy (PE) of the glider is mgh, where m is the mass of the glider, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the glider was released.

3. The final kinetic energy (KE) of the glider is 1/2mv², where v is the final speed of the glider.

4. Since the track is frictionless, the initial potential energy equals the final kinetic energy, so mgh = 1/2mv².

5. We can cancel out the mass m from both sides of the equation, which gives us gh = 1/2v².

6. We can solve this equation for h, the height from which the glider was released: h = v² / (2*g).

7. The height h can also be expressed in terms of the distance d and the angle of inclination θ of the track: h = d*sin(θ).

8. Therefore, we can set v² / (2g) = dsin(θ) and solve for θ.

9. Rearranging the equation gives us θ = arcsin(v² / (2gd)).

10. Substituting the given values into this equation gives us θ = arcsin((0.233 m/s)² / (29.8 m/s²0.605 m)).

11. Calculating this gives us the angle of inclination of the track in radians. To convert this to degrees, multiply by 180/π.

Please note that the mass of the glider was not given in the problem, but it cancels out in the equations, so it does not affect the final answer.