91Ó°ÊÓ

Mass of the block is M.

Acceleration is $\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$.

In this problem the cord is doing work on the block so that it does not

undergo free fall.

Use F to denote the magnitude of the force of the cord on the block. This force is upward, opposite to the force of gravity (which has magnitude$F_g=Mg$ ) to prevent the block from undergoing free fall.

The downward acceleration is$\stackrel{→}{a}=\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Taking the downward direction to be positive, then Newton’s second law yields

$\begin{array}{rcl}\stackrel{→}{F_{net}}& =& m\stackrel{→}{a}\\ & ⇒& Mg-F=M\left(\frac{g}{4}\right)\\ F& =& \frac{3Mg}{4}\\ & & \end{array}$

The force is in the opposite direction of the displacement.

On the other hand, the force of gravity is,

$F_g=Mg$

This force is in the same direction to the displacement.

Since the displacement is downward, the work done by the cord’s force is,

$\begin{array}{rcl}{W}_F& =& -Fd\\ & =& -\frac{3}{4}Mgd\\ & & \end{array}$

Similarly, the work done by the force of gravity is,

$\begin{array}{rcl}{W}_g& =& F_{g}d\\ & =& Mgd\\ & & \end{array}$

The total work done on the block is simply the sum of the two works:

$\begin{array}{rcl}{W}_{net}& =& W_F+W_g\\ & =& -\frac{3}{4}Mgd+Mgd\\ & =& \frac{1}{4}Mgd\\ & & \end{array}$

Since the block starts from rest, this$\left(\frac{Mgd}{4}\right)$is the block’s kinetic energy k at the moment it has descended the distance d .

Write the kinetic energy equation as below.

$K=\frac{1}{2}M{v}^2$

Rearrange the above equation for the speed as follow.

$\begin{array}{rcl}v& =& \sqrt{\frac{2K}{M}}\\ & =& \sqrt{\frac{2\left({\displaystyle \raisebox{1ex}{${\displaystyle Mgd}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 4}$}\right.}\right)}{{\displaystyle M}}}\\ & =& \sqrt{\frac{gd}{2}}\\ & & \\ & & \end{array}$

At the moment the block has descended the distance d.