Question

A 1.0-kg object moving in a certain direction has a kinetic energy of 2.0 J. It hits a wall and comes back with half its original speed. What is the kinetic energy of this object at this point?

Answer 1

Lets calculate the initial speed of the block. We know that kinetic energy is given by:

`K_0 =(mv_0^2)/(2)`

Solving for v₀:

`v_0 = \sqrt{ (2K_0)/(m)} = \sqrt{ (2(2\;J))/(1\;kg)} = 2\;m/s`

If the speed after it hits a wall is half its original speed then:

v = v₀/2 = 2 m/s / 2 = 1 m/s

Then the kinetic energy at this point is:

`K =(mv^2)/(2) = ((1\;kg)(1\;m/s)^2)/(2) = (1)/(2)\;J = 0.5\;J`

The inetic energy of this object at this point is 0.5 J.