An object slides down a frictionless 33 degree incline whose vertical height is 37.0 cm. How fast is it going in meters/second when it reaches the bottom?
To determine the speed of the object at the bottom of the incline, we can use the law of conservation of energy, which states that the initial potential energy of the object will be converted into kinetic energy as it reaches the bottom.
The potential energy (PE) of the object at the top of the incline can be calculated using the formula:
PE = m * g * h,
where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height of the incline (37.0 cm or 0.37 m).
Since the incline is frictionless, all of the potential energy is converted into kinetic energy at the bottom. Therefore, we can equate the two energies:
PE = KE,
m * g * h = 0.5 * m * v^2,
where v is the final velocity of the object at the bottom.
Simplifying the equation, we have:
g * h = 0.5 * v^2,
v^2 = (2 * g * h),
v = √(2 * g * h).
Now we can substitute the known values:
v = √(2 * 9.8 m/s^2 * 0.37 m),
v ≈ √(7.252 m^2/s^2),
v ≈ 2.69 m/s.
Therefore, the object is going approximately 2.69 meters per second when it reaches the bottom of the incline.