Question I

I) A cord is used to vertically lower an initially stationary block of mass m at a constant downward acceleration of g/4. When the block has fallen a distance d, find (a) the work done by the cord's force on the block (7 pts), (b) the work done by the weight of the block (3 pts), (c) the kinetic energy of the block (7 pts), and (d) the speed of the block (3 pts)

Solution:
a)
If we sum the forces acting on our block, we find
$T - mg = m (\frac{-g}{4})$
So the tension is given by $T = \frac{3mg}{4}$.
The work done by the tension is the force times the distance, or
W_T = T(-d)
$W_{T} = \frac{3mg}{4} (-d)$
$W_{T} = -\frac{3}{4}mgd$

b)
The work done by the weight of the block is given by
W_g = (-mg)(-d)
W_g = mgd

c)
The energy left over is
KE = W_g + W_T
$KE = mgd - \frac{3}{4}mgd ~=~ \frac{1}{4}mgd$

d)
So, the velocity is given by
$KE = \frac{1}{2} mv^{2}$
$v^{2} = \frac{1}{2} gd$
$v = \sqrt{\frac{gd}{2}}$