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. 50.9 J
C. 0 J
D. 44.1 J

Answer 1

The answer is D. $E_k = 44.1\ J$

To answer this problem we must make an energy balance. Bear in mind that mechanical energy is preserved

Call instant (1) at the moment when the block is at the top of the inclined plane at a height of 3 m.

Let's call instant (2) the moment in which the block has descended by the inclined plane and its height is 0 m.

Then:

$E_{(1)} = E_{(2)}$

At the instant (1) the block is stopped, this means that its kinetic energy = 0 and its gravitational potential energy is maximum.

$E_{(1)} = mgh$

At the instant (2) the block is in motion and has reached the end of the inclined plane (height = 0 m)

$E_{(2)} = \frac{1}{2}mv^2 = E_k$

Where $E_k$ = kinetic energy

So:

$mgh = E_k$

$E_k = 1.5(9.8)(3)$

$E_k = 44.1\ J$