Lets calculate the initial speed of the block. We know that kinetic energy is given by:
[tex]K_0 =\dfrac{mv_0^2}{2}[/tex]
Solving for v₀:
[tex]v_0 = \sqrt{ \dfrac{2K_0}{m}} = \sqrt{ \dfrac{2(2\;J)}{1\;kg}} = 2\;m/s[/tex]
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:
[tex]K =\dfrac{mv^2}{2} = \dfrac{(1\;kg)(1\;m/s)^2}{2} = \dfrac{1}{2}\;J = 0.5\;J[/tex]
The inetic energy of this object at this point is 0.5 J.
