Final answer:
The velocity of a 1 kg block sliding down a 2.5 meter high frictionless inclined plane is found using conservation of energy, resulting in a final velocity of 7 m/s at the bottom of the incline.
Step-by-step explanation:
To determine the velocity of a 1 kilogram block sliding down a frictionless inclined plane which is 2.5 meters high, we can use the principles of conservation of energy or the equations of motion. Without friction or air resistance to dissipate energy, the potential energy of the block at the top of the incline will convert entirely into kinetic energy at the bottom of the incline, assuming the block starts from rest.
Using the conservation of energy approach, the potential energy (PE) at the top is equal to the kinetic energy (KE) at the bottom:
PE = KE,
mgh = (1/2)mv²,m is the mass of the block, g is the acceleration due to gravity (9.8 m/s²), h is the height of the incline, and v is the final velocity. Solving for v, we get:
v = √(2gh).
Substituting the given values (m=1 kg, g=9.8 m/s², and h=2.5 m), we get:
v = √(2 * 9.8 m/s² * 2.5 m) = √(49 m²/s²) = 7 m/s.
Therefore, the velocity of the block when it reaches the bottom of the plane is 7 m/s.