A 90.0 kgmail bag hangs by a vertical rope 3.5 mlong. A postal worker then displaces the bag to a position 2.0 msideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

The given data can be listed below as:

• The mass of the mail bag is, $m=90.0\mathrm{kg}$.
• The length of the vertical rope is, $I=3.5\mathrm{m}$.
• The bag is displaced to a distance of $s=2.0\mathrm{m}$.

The work done is described as the product of the force exerted on an object and the distance moved by that object. The work done on an object is also the energy transferred from that object.

The free body diagram of the mail bag has been drawn below:

From the free body diagram, it can be identified that the summation of the forces in the x and in the y direction is zero.

The equation of the summation of the forces in the x direction is expressed as:

$$\begin{gathered} \sum _{}{\mathrm{F}}_{\mathrm{x}}=0 \\ \mathrm{Tsin\theta }=0 \end{gathered}$$

Here,$\sum _{}{\mathrm{F}}_{\mathrm{x}}$ is the summation of the forces in the x direction, T is the tension exerted on the bag and$\theta$ is the angle subtended by the bag.

The equation of the summation of the forces in the y direction is expressed as:

$$\begin{gathered} \sum _{}{\mathrm{F}}_y=0 \\ T\cos \theta -mg=0 \end{gathered}$$

Here,$\sum _{}{\mathrm{F}}_y$ is the summation of the forces in the y direction, T is the tension exerted on the bag, is the mass of the mail bag, is the acceleration due to gravity and$\theta$ is the angle subtended by the bag.

The equation of the angle subtended by the bag is expressed as:

$\theta ={\sin }^{-1}\frac{s}{I}$

Here, s is the displacement of the bag and$I$ is the length of the rope.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{\theta }={\sin }^{-1}\frac{2\mathrm{m}}{3.5\mathrm{m}} \\ ={\sin }^{-1}\left(0.57\right) \\ =34.85° \end{gathered}$$

The equation of the horizontal force is expressed as:

$F=mg\tan \theta$

Here, $F$is the horizontal force.

Substitute the values in the above equation.

role="math" localid="1667897323440" $$\begin{gathered} \mathrm{F}=\left(90\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\tan 34.85° \\ =\left(0.108\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2\right)0.69 \\ =614.15\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2} \\ =614.15\mathrm{N} \end{gathered}$$

Thus, the horizontal force necessary to hold the bag in the new position is 614.15 N.

As there is no displacement happened of the rope, then the initial and the final displacement of the rope is zero. Hence, the work done by the rope is zero.

Thus, the work done by the rope is zero.

The equation of the work done is expressed as:

$W=\mathrm{mgs}$

Here, W is the work done, m is the mass of the mail bag, g is the acceleration due to gravity and s is the displacement of the bag.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{W}=\left(90\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{m}\right) \\ =\left(822\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{m}\right) \\ =1764\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2} \\ =1764\mathrm{J} \end{gathered}$$

Thus, the work done by the worker is 1764 J .