A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of $\mathbf{10}\mathbf{.}\mathbf{0}\mathit{m}$: (a) the initially stationary spelunker is accelerated to a speed of $\mathbf{5}\mathbf{.}\mathbf{00}\mathit{m}\mathbf{/}\mathit{s}$; (b) he is then lifted at the constant speed of$\mathbf{5}\mathbf{.}\mathbf{00}\mathit{m}\mathbf{/}\mathit{s}$ ; (c) finally he is decelerated to zero speed. How much work is done on the $\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{}}\mathit{k}\mathit{g}\mathbf{}$rescuee by the force lifting him during each stage?

1. The mass of the rescuee is,$m=80\text{kg}$
2. Height is,$h=10.0\text{m}$
3. Velocity is$v=5.0\text{m/s}$

By usingtheconcept of change in kinetic as well as potential energy, we can find the work done during each step.

Formula:

$\Delta KE=\frac{1}{2}m{v}^2$

Here,g is the gravitational acceleration whose value is$9.8{\text{m/s}}^{\text{2}}$

Potential energycan be calculated as,

$PE=mgh$ (1)

Substituting the values in the above expression, and we get,

$\begin{array}{rcl}PE& =& \left(80\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(10\text{m}\right)\\ & =& 7840·\left(1\text{kg}\times 1{\text{m/s}}^{\text{2}}\times 1\text{m}\times \frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\\ PE& =& 7840\text{J}\\ & & \end{array}$ (2)

Change Kinetic energy can be calculated as,
$\Delta KE=\frac{1}{2}m{v}^{\text{2}}$ (3)

Substituting the values in the above expression, and we get,

$$\begin{gathered} \Delta KE=\frac{1}{2}\left(80\text{kg}\right){\left(5\text{m/s}\right)}^2 \\ =1000·\left(1\text{kg}\times 1{\text{m}}^2{\text{/s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right) \\ =1000\text{J} \end{gathered}$$

From the work-energy theorem, the total change in energy and the work done on the system will be equal; then, we can write,

$W_1=\Delta KE+\Delta PE$

Substituting the values in the above expression, and we get,

$\begin{array}{rcl}{W}_{\text{1}}& =& 1000\text{J}+7840\text{J}\\ W_{\text{1}}& =& 8840\text{J}\\ & =& 8.84\times {10}^3\text{J}\\ & & \end{array}$

Thus, work done when spelunker is accelerated to a speed of $5.00\text{m/s}$is $8.84\times {10}^3\text{J}$.

In this case, the speed is constant.

So total energy will be the potential energy, and kinetic energy will remain the same.

From equation (2), the potential energy will be,

$PE=7840\text{J}$

Thework done can be written as,

$\begin{array}{rcl}{W}_{\text{2}}& =& PE\\ & =& 7840\text{J}\\ & =& 7.84\times {10}^3\text{J}\\ & & \end{array}$

Thus, the work done when he is lifted with a constant speed of $5.00\text{m/s}$is $7.84\times {10}^3\text{J}$.

In this case, the velocity decreases to zero from 5 m/s.

Then from equation 3, the decrease in kinetic energy can be calculated as,

$\begin{array}{rcl}\Delta KE& =& 0-\frac{1}{2}\left(80\text{kg}\right){\left(5\text{m/s}\right)}^2\\ & =& -1000·\left(1\text{kg}\times 1{\text{m}}^2{\text{/s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\\ & =& -1000\text{J}\\ & & \end{array}$

From equation (2), potential energy is,

$PE=7840\text{J}$

From the work-energy theorem, the total change in energy and the work done on the system will be equal; then, we can write,

$W_3=\Delta KE+\Delta PE$

Substituting the values in the above expression, and we get,

$\begin{array}{rcl}{W}_3& =& 7840\text{J}-1000\text{J}\\ W_{\text{3}}& =& 6840\text{J}\\ & =& 6.84\times {10}^3\text{J}\\ & & \end{array}$

Thus, work done when he is decelerated to zero speed $6.84\times {10}^3\text{J}$.