Answer
The work done on the block by friction is -0.230 J.
Work Step by Step
We can find the speed at point A.
$\frac{mv_A^2}{R} = \sum F$
$\frac{mv_A^2}{R} = F_N-mg$
$v_A^2 = \frac{R~(F_N-mg)}{m}$
$v_A = \sqrt{\frac{R~(F_N-mg)}{m}}$
$v_A = \sqrt{\frac{(0.500~m)[3.95~N-(0.0400~kg)(9.80~m/s^2)]}{0.0400~kg}}$
$v_A = 6.67~m/s$
We can find the speed at point B.
$\frac{mv_B^2}{R} = \sum F$
$\frac{mv_B^2}{R} = F_N+mg$
$v_B^2 = \frac{R~(F_N+mg)}{m}$
$v_B = \sqrt{\frac{R~(F_N+mg)}{m}}$
$v_B = \sqrt{\frac{(0.500~m)[0.680~N+(0.0400~kg)(9.80~m/s^2)]}{0.0400~kg}}$
$v_B = 3.66~m/s$
$K_1+U_1+W=K_2+U_2$
$W = K_2+U_2-K_1-U_1$
$W = \frac{1}{2}mv_B^2+2mgR-\frac{1}{2}mv_A^2-0$
$W = \frac{1}{2}(0.0400~kg)(3.66~m/s)^2+(2)(0.0400~kg)(9.80~m/s^2)(0.500~m)-\frac{1}{2}(0.0400~kg)(6.67~m/s)^2$
$W = -0.230~J$
The work done on the block by friction is -0.230 J.
