solution

(a) So how do we use the work energy theorem? Well first we calculate the work. The initial position $x_i = -.02 m$ and the final position $x_f ~=~ 0$. From eq. 1.7 we see that

$$W ~=~ -{1\over 2}k (0- x_i^2) ~=~ {1\over 2}k x_i^2$$ (1.8)

What about the change in kinetic energy? The initial kinetic energy is zero so $\Delta K ~=~ {1\over 2}m v_f^2$.

Equating $\Delta K$ to $W$ we have

$${1\over 2}m v_f^2 ~=~ {1\over 2}k x_i^2$$ (1.9)

which simplifies to

$$v_f ~=~ \sqrt{k\over m} \vert x_i\vert ~=~ \sqrt{ 1000 Nm/1.6 kg} .02 m ~=~ 0.5 m/s ~.$$ (1.10)

(b) In this case we have an additional force acting on the block, It is a constant force and opposed the direction of motion. Therefore the work done on the block due to this frictional force is $W_f ~=~ f \Delta x$ where $f ~=~ 4N$.

So get the net work on the block we have to add in this new force:

$$W_{net} ~=~ {1\over 2}k x_i^2 - f\Delta x$$ (1.11)

Again, by the work energy theorem, this must be equal to the change in kinetic energy, so we have

$${1\over 2}k x_i^2 - f\Delta x = {1\over 2}m v_f^2$$ (1.12)

Solving for $v_f$, we have

$$v_f = \sqrt{{k\over m} x_i^2 - 2{f\Delta x\over m}} ~=~ .39 m/s$$ (1.13)