Answer
(a) $2.55m/s^2$ opposite the direction of motion.
(b) $4.26m/s$
(c) Please see the work below.
Work Step by Step
(a) We can find the required acceleration as follows:
$\Sigma F_x=\mu_k N$
$\implies ma=-\mu_kmg$
$\implies a=-\mu_k g$
We plug in the known values to obtain:
$a=-(0.260)(9.81)$
$a=-2.55m/s^2$
The negative sign shows that it is opposite to the direction of motion.
(b) We can find the final speed as
$v_f^2=v_i^2+2ad$
$\implies v_f=\sqrt{v_i^2+2ad}$
We plug in the known values to obtain:
$v_f=\sqrt{(4.33)^2+2(-2.55)(0.125)}$
$v_f=4.26m/s$
(c) We know that
$v_f^2=v_i^2+2ad$
$\implies v_f^2=v_i^2+2(\frac{F}{m})d$
Multiplying both sides by $\frac{1}{2}m$, we obtain:
$\frac{1}{2}mv_f^2=\frac{1}{2}mv_i^2+Fd$
This can be rearranged as
$Fd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
$\implies -Fd=-(\mu_k mg)d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$
