Answer
(a) The work done was -6.56 J.
(b) v = 0.750 m/s
(c) v = 1.60 m/s
Work Step by Step
(a) We can find the kinetic energy at point A.
$KE_A = \frac{1}{2}mv^2 = \frac{1}{2}(1.50~kg)(3.21~m/s)^2$
$KE_A = 7.728~J$
We can find the kinetic energy at point B.
$KE_B = \frac{1}{2}mv^2 = \frac{1}{2}(1.50~kg)(1.25~m/s)^2$
$KE_B = 1.172~J$
We can find the change in kinetic energy.
$\Delta KE = KE_B-KE_A = 1.172~J - 7.728~J$
$\Delta KE = -6.56~J$
The work done was -6.56 J.
(b) We can find the kinetic energy at point C.
$KE_C - KE_B = -0.750~J$
$KE_C = KE_B-0.750~J$
$KE_C = 1.172~J-0.750~J$
$KE_C = 0.422~J$
We can find the speed $v$ at point C.
$ \frac{1}{2}mv^2 = 0.422~J$
$v^2 = \frac{0.844~J}{1.50~kg}$
$v = \sqrt{\frac{0.844~J}{1.50~kg}}$
$v = 0.750~m/s$
(c) We can find the kinetic energy at point C.
$KE_C - KE_B = 0.750~J$
$KE_C = KE_B+0.750~J$
$KE_C = 1.172~J+0.750~J$
$KE_C = 1.922~J$
We can find the speed $v$ at point C.
$ \frac{1}{2}mv^2 = 1.922~J$
$v^2 = \frac{3.844~J}{1.50~kg}$
$v = \sqrt{\frac{0.844~J}{1.50~kg}}$
$v = 1.60~m/s$