Question

A 1.50 kgwater balloon is shot straight up with an initial speed of300 m/s (a) What is the kinetic energy of the balloon just as it is launched? (b) How much work does the gravitational force do on the balloon during the balloon’s full ascent? (c) What is the change in the gravitational potential energy of the balloon-Earth system during the full ascent? (d) If the gravitational potential energy is taken to be zero at the launch point, what is its value when the balloon reaches its maximum height? (e) If instead, the gravitational potential energy is taken to be zero at the maximum height, what is its value at the launch point? (f) What is the maximum height?

Short answer

1. The kinetic energy of the balloon at the launch will be 6.75 J .
2. Work done by the gravitational force on the ball during the balloon’s full ascent will be -6.75 J .
3. Change on the balloon-Earth system during full ascent will be 6.75 J .
4. Gravitational potential when the balloon reaches its maximum height is 6.75 J
5. Gravitational potential at the launch point when its value at a maximum point is zero will be -6.75 J.
6. Maximum height reached by the balloon will be 0.46 m.

Step-by-step solution

The given data

The mass of the water balloon is, m = 1.50 kg

The initial speed of the water balloon is,$v_i=3m/s$

Understanding the concept of energy

As the ball is launched from its initial position, using the formula for kinetic energy, we can find its kinetic energy when it is launched. As it reaches its full ascent, its kinetic energy will be converted to potential energy which will be called the work done by the gravitational force on the balloon. As we know the gravitational potential at the initial and final position, we can find the change in the gravitational potential. Using the equation for kinematics we can find the maximum height reached by the balloon.

Formulae:

The potential energy at a height, PE = mgh (1)

The kinetic energy of the body, $\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$ (2)

The change in the potential energy, $∆PE=P{E}_{final}-P{E}_{initial}$ (3)

The work done by an applied force, $W=Force\times displacement$ (4)

The third equation of the kinematic equation,

$v_f^2=v_i^2+2ax$ (5)

a) Calculation of kinetic energy at the launch

Kinetic energy at the launch point can be given using equation (2) as follows:

$$\begin{gathered} K{E}_{initial}=\frac{1}{2}\times 1.50kg\times {\left(3.00m/s\right)}^2 \\ =6.75kg·m^2/s^2\left(\frac{1J}{1kg·m^2/s^2}\right) \\ =6.75J \end{gathered}$$

Hence, the value of the kinetic energy is 6.75 J.

b) Calculation of work done by the gravitational force

When it will reach the maximum height, its final velocity ($v_f=0$) will be zero, thus the distance traveled by the balloon can be given using equation (5):

$$\begin{gathered} 0m/s={\left(3.0m/s\right)}^2-2\times 9.8m/s^2\times \left(-x\right) \\ 9m^2/s^2=-19.6m/s^2\times x \\ x=-\frac{9m^2/s^2}{19.6m/s^2} \\ x=-0.46m \end{gathered}$$

The force of gravitation on the balloon can be given as:

$$\begin{gathered} F_{Gravitation}=mg \\ =1.50kg\times 9.8m/s^2 \\ =14.7kg·m/s^2\left(\frac{1N}{1kg·m/s^2}\right) \\ =14.7N \end{gathered}$$

Now, the work done by the force can be given using equation (4) as:

$$\begin{gathered} W=14.7N\times \left(-0.46m\right) \\ =-6.75N·m\left(\frac{1J}{1N·m}\right) \\ =-6.75J \end{gathered}$$

Hence, the value of the work done is -6.75 J.

c) Calculation of the change in the potential energy of the system

When the ball is at the launch position, h = 0 , thus the initial potential energy using equation (1) is given as:

${\mathrm{PE}}_{\mathrm{initial}}=0J$

When the ball is at maximum height, h = 0.46 m, thus the final potential energy using equation (1) is given as:

$$\begin{gathered} P{E}_{final}=1.5kg\times 9.8m/s^2\times 0.46m \\ =6.75kg·m^2/s^2\left(\frac{1J}{1kg·m^2/s^2}\right) \\ =6.75J \end{gathered}$$

Now, the change in potential energy is given using equation (3) as follows:

$$\begin{gathered} ∆PE=P{E}_{final}-P{E}_{initial} \\ =6.75J-0J \\ =6.75J \end{gathered}$$

Hence, the value of the change in the energy is 6.75 J.

d) Calculation of the potential at maximum height

From the above calculation, it can be seen that the potential energy at maximum height, h = 0.46 m is 6.75 J.

e) Calculation of the potential at the launch point when the potential at maximum is taken to be zero

Now at the maximum height, the gravitational potential is$P{E}_{final}=0J$

When the ball is at its initial position, its distance from the final position will be h = 0.46 m

From part (d), the change in the potential energy can be given as:

$∆PE=6.75J$

So, the initial potential energy is given using equation (3):

$$\begin{gathered} P{E}_{initial}=∆P{E}_{final}-∆PE \\ =0J-6.75J \\ =-6.75J \end{gathered}$$

Hence, the value of the potential at the launch point is -6.75 J

f) Calculation of the maximum height

When it will reach the maximum height, its velocity will be zero, hence, from the part (b) calculations, the maximum height of the balloon is 0.46 m.