A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block starts at a height of 2 m, how

Question

 how much kinetic energy does the block have when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2.

Answer 1

There are a few different ways to solve this problem.  Since there is no friction, mechanical energy is conserved, hence the energy conservation method may be the easiest to use in this problem.

 

U₁ = gravitational potential energy at the top of the plane = mgh

K₁ = kinetic energy at the top of the plane = 0 (starts from rest)

U₂ = gravitational potential energy at the bottom of the plane = 0 (height is zero)
K₂ = kinetic energy at the bottom of the plane = (1/2)mv²

 

Conservation of mechanical energy:

 

U₁ + K₁ = U₂ + K₂

 

mgh + 0 = 0 + K₂

 

K₂ = mgh = (10 kg)(9.8 m/s²)(2 m) = 196 J

 

Note the inclination angle of the plane does not appear in the solution.