Question

a block of mass 1.5 kg slides down an inclined plane that has an angle of 15°. if the inclined plane has no friction and the block starts at a height of 3 m, how much kinetic energy does the block have when it reaches the bottom? acceleration due to gravity is g = 9.8 m/s².
a. 6.8 j
b. 0 j
c. 44.1 j
d. 50.9 j

Explanation:

Step1: Apply the law of conservation of mechanical energy

Since there is no friction, the initial potential energy \(U = mgh\) is converted entirely into kinetic energy \(K\) at the bottom.

Step2: Calculate the initial potential energy

Given \(m = 1.5\space kg\), \(h=3\space m\), \(g = 9.8\space m/s^{2}\).
Using the formula \(U=mgh\), we substitute the values:
\(U=(1.5\space kg)\times(9.8\space m/s^{2})\times(3\space m)\)
\(U = 44.1\space J\)
Since \(K = U\) (conservation of energy, no non - conservative forces), the kinetic energy at the bottom is \(44.1\space J\)

Answer:

C. 44.1 J