Chapter 4, Problem 80A

(a)

To determine

Theacceleration of the balloon and the instruments.

Answer to Problem 80A

  $2.45{\text{m/s}}^2$

Explanation of Solution

Given:

The mass of the instruments in a weather balloon is $m=8\text{kg}$ .

The upward force exerted by the balloon on the instrument is $F_{\text{applied}}=98\text{N}$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Here, $m$ is the mass and $a$ is the acceleration of the object.

Calculation:

Consider the acceleration due to gravity as $-9.80\text{m/s}$ .

The negative sign indicates the movement of the balloon against the gravity.

Find the net force acting on the assembly of the balloon and the instruments as follows:

  $\begin{array}{l}{F}_{\text{net}}=F_{\text{applied}}+F_{\text{g}}\\ F_{\text{net}}=F_{\text{applied}}+m\left(-g\right)\\ F_{\text{net}}=98\text{N}+8.0\text{kg}\times \left(-9.80{\text{m/s}}^2\right)\\ F_{\text{net}}=98\text{N}-78.4\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ F_{\text{net}}=98\text{N}-78.4\text{N}\\ F_{\text{net}}=19.6\text{N}\end{array}$

Find the acceleration of the balloon and the instruments as follows:

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ a=\frac{19.6\text{N}\times \left(\frac{1\text{kg}\cdot {\text{m/s}}^2}{1\text{N}}\right)}{8\text{kg}}\\ a=\frac{19.6\text{kg}\cdot {\text{m/s}}^2}{8\text{kg}}\\ a=2.45{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the acceleration of the balloon and the instruments is $2.45{\text{m/s}}^2$ .

(b)

To determine

The velocity of the instruments when the balloon is accelerated for $\text{10}\text{s}$ and the instrument is released.

Answer to Problem 80A

The velocity of the instruments is $24.5\text{m/s}$ upwards.

Explanation of Solution

Given:

The mass of the instruments in a weather balloon is $m=8\text{kg}$ .

The upward force exerted by the balloon on the instrument is $F_{\text{applied}}=98\text{N}$ .

Formula used:

Show the expression for the acceleration of the object as follows:

  $a=\frac{v}{t}$

Here, $v$ is the velocity of the object and $t$ is the time of acceleration of the object.

Calculation:

Refer to part (a).

The acceleration of the balloon and the instrument is $a=2.45{\text{m/s}}^2$ .

Find the velocity of the instrument as follows:

  $\begin{array}{c}v=at\\ =2.45{\text{m/s}}^2\times 10\text{s}\\ =\text{24}\text{.5}\text{m/s}\left(\text{upward}\right)\end{array}$

Conclusion:

Thus, the velocity of the instrument is $\text{24}\text{.5}\text{m/s}\left(\text{upward}\right)$ .

(c)

To determine

The net force acting on the instrument after release.

Answer to Problem 80A

The net force acting on the instrument after release is $-19.6\text{N}$ downwards.

Explanation of Solution

Given info:

The mass of the instruments in a weather balloon is $m=8\text{kg}$ .

The upward force exerted by the balloon on the instrument is $F_{\text{applied}}=98\text{N}$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Here, $m$ is the mass and $a$ is the acceleration of the object.

Calculation:

Find the net force acting on the instruments after the release as follows:

  $\begin{array}{l}{F}_{\text{net}}=-\left\{F_{\text{applied}}+F_{\text{g}}\right\}\\ F_{\text{net}}=-\left\{F_{\text{applied}}+m\left(-g\right)\right\}\\ F_{\text{net}}=-\left\{98\text{N}+8.0\text{kg}\times \left(-9.80{\text{m/s}}^2\right)\right\}\\ F_{\text{net}}=-\left\{98\text{N}-78.4\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\right\}\\ F_{\text{net}}=-98\text{N}+78.4\text{N}\\ F_{\text{net}}=-19.6\text{N}\end{array}$

The negative sign indicates that the net force acting on the instruments after the release acts in a downward direction.

Conclusion:

Thus, the net force acting on the instrument after release is $-19.6\text{N}$ (downwards).

(d)

To determine

Thetime when the direction of the velocity becomes downward.

Answer to Problem 80A

The direction of the velocity becomes downward after $t=2.5\text{s}$ of release.

Explanation of Solution

Given:

The mass of the instruments in a weather balloon is $m=8\text{kg}$ .

The upward force exerted by the balloon on the instrument is $F_{\text{applied}}=98\text{N}$ .

Formula used:

Equation of Motion:

Show the expression for the equation of motion as follows:

  $v_{\text{f}}=v_{\text{i}}+gt$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, $a$ is the acceleration, $g$ is the acceleration due to gravity, and $t$ is the time.

Calculation:

Consider the acceleration due to the gravity of the Earth is $-9.80{\text{m/s}}^2$ .

The final velocity of the instrument is $v_{\text{f}}=0\text{m/s}$ .

The initial velocity of the instrument is

Find the time taken for the velocity to change the direction as follows:

  $\begin{array}{l}0=24.5-9.80t\\ t=\frac{24.5}{9.80}\\ t=2.5\text{s}\end{array}$

Conclusion:

The direction of the velocity becomes downward after $t=2.5\text{s}$ of release.