A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of$\raisebox{1ex}{$\mathbf{g}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{4}$}\right.$ .When the block has fallen a distance d , find

(a) the work done by the cord’s force on the block,

(b) the work done by the gravitational force on the block,

(c) the kinetic energy of the block, and

(d) the speed of the block.

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$\overrightarrow{a}=\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

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

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

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.