Chapter 9, Problem 10PQ

A helicopter rescues a trapped person of mass m = 65.0 kg from a flooded river by lifting the person vertically upward using a winch and rope. The person is pulled 12.0 m into the helicopter with a constant force that is 15% greater than the person’s weight. a. Find the work done by each of the forces acting on the person. b. Assuming the survivor starts from rest, determine his speed upon reaching the helicopter.

To determine

The work done by each of the forces acting on the person.

Answer to Problem 10PQ

The work done by gravitational force on the person is $-7.65\times {10}^3\text{N}\cdot \text{m}$ and work done by tension on the string is $+8.80\times {10}^3\text{N}\cdot \text{m}$ .

Explanation of Solution

In this problem, there are two forces acting on the person which are the gravitational force of attraction or weight acting downward and tension in the rope acting upward.

Write the expression for the weight of the person.

  $F_g=mg$                                                                                                                    (I)

Here, $F_g$ is the weight of the person, $m$ is the mass of the person and $g$ is the acceleration due to gravity.

It is given that tension is $15\%$ greater than weight of the person.

Write the expression for the tension in the rope.

  $\begin{array}{c}{F}_T=F_g+\frac{15}{100}{F}_g\\ =1.15F_g\end{array}$                                                                                                      (II)

Here, $F_T$ is the tension in the string.

Helicopter is lifting the person vertically upward. The gravitational force on person is antiparallel to displacement of the person. The tension is parallel to displacement of the person. This implies that work done by gravitational force is negative and that by tension is positive.

Write the expression for the work done by a force.

  $W=F\Delta r\cos \theta$                                                                                                       (III)

Here, $W$ is the work done, $F$ is the force, $\Delta r$ is the displacement and $\theta$ is the angle between force and displacement.

Using equation (III),write the expression for the work done by gravitational force on person.

  $\begin{array}{c}{W}_g=F_g\Delta r\cos 180°\\ =-F_g\Delta r\end{array}$                                                                                                 (IV)

Here. $W_g$ is the work done by gravitational force of attraction.

Using equation (III),write the expression for the work done by tension in rope on person.

  $W_T=F_T\Delta r$                                                                                                               (V)

Here. $W_T$ is the work done by tension in the string.

Conclusion:

Substitute $65.0\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (I) to get $F_g$.

  $\begin{array}{c}{F}_g=\left(65.0\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ =638\text{N}\end{array}$

Substitute $638\text{N}$ for $F_g$ in equation (II) to get $F_T$.

  $\begin{array}{c}{F}_T=1.15\left(638\text{N}\right)\\ =733.7\text{N}\end{array}$

Substitute $638\text{N}$ for $F_g$ and $12.0\text{m}$ for $\Delta r$ in (IV) to get $W_g$.

  $\begin{array}{c}{W}_g=-\left(638\text{N}\right)\left(12.0\text{m}\right)\\ =-7656\text{J}\\ =-\text{7}\text{.65}\times {\text{10}}^3\text{N}\cdot \text{m}\end{array}$

Substitute $733.7\text{N}$ for $F_g$ and $12.0\text{m}$ for $\Delta r$ in (V) to get $W_T$.

  $\begin{array}{c}{W}_T=\left(733.7\text{N}\right)\left(12.0\text{m}\right)\\ =8.80\times {10}^3\text{N}\cdot \text{m}\end{array}$

Therefore, the work done by gravitational force on the person is $-7.65\times {10}^3\text{N}\cdot \text{m}$ and work done by tension on the string is $+8.80\times {10}^3\text{N}\cdot \text{m}$ .

To determine

The speed of survivor upon reaching the helicopter assuming he starts from rest.

Answer to Problem 10PQ

The speed of survivor upon reaching the helicopter assuming he starts from rest is $5.95\text{m}/\text{s}$ .

Explanation of Solution

Write the equation for the change in kinetic energy of the person using work energy theorem.

  $\Delta K=W_{\text{tot}}$                                                                                                                (VI)

Here, $\Delta K$ is the change in kinetic energy of the person and $W_{\text{tot}}$ is the total work done on the person by all forces.

Write the expression for the total work done.

  $W_{\text{tot}}=W_g+W_T$                                                                                                        (VII)

Here, $W_{\text{tot}}$ is the total work done on the person.

Write the expression for change in kinetic energy.

  $\Delta K=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$                                                                                             (VIII)

Here, $v_i$ is the initial velocity of the person and $v_f$ is the final velocity of the person.

Substitute equation (VIII) and (VII) in equation (VI) to expand the equation.

  $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=W_g+W_T$

Substitute $0\text{m}/\text{s}$ for $v_i$ in above equation to get $v_f$.

  $\begin{array}{c}\frac{1}{2}{v}_f^2-\frac{1}{2}m\left(0\text{m}/\text{s}\right)=W_g+W_T\\ v_f=\sqrt{2\left(\frac{W_g+W_T}{m}\right)}\end{array}$                                                                       (IX)

Conclusion:

Substitute $-7.65\times {10}^3\text{N}\cdot \text{m}$ for $W_g$, $+8.80\times {10}^3\text{N}\cdot \text{m}$ for $W_T$ and $65.0\text{kg}$ for $m$ in equation (IX) to get $v_f$ .

  $\begin{array}{c}{v}_f=\sqrt{2\left(\frac{-7.65\times {10}^3\text{N}\cdot \text{m}+8.80\times {10}^3\text{N}\cdot \text{m}}{65.0\text{kg}}\right)}\\ =5.95\text{m}/\text{s}\end{array}$

Therefore, The speed of survivor upon reaching the helicopter assuming he starts from rest is $5.95\text{m}/\text{s}$ .