A 2.50 kgbeverage can is thrown directly downward from a height of 4.00m, with an initial speed of 3.00 m/s. The air drag on the can is negligible. What is the kinetic energy of the can (a) as it reaches the ground at the end of its fall and (b) when it is halfway to the ground? What are (c) the kinetic energy of the can and (d) the gravitational potential energy of the can-Energy system 0.200 sbefore they can reach the ground? For the latter, take the reference pointy = 0 to be at the ground.

Mass of the beverage can is,$\mathrm{m}=2.50\mathrm{kg}$

Height from which can is thrown is, $\mathrm{h}=4.00\mathrm{m}$

The initial speed of the bottle can is,${\mathrm{v}}_{\mathrm{i}}=3.00\mathrm{m}/\mathrm{s}$

Time before reaching the ground is,${\mathrm{t}}_{\mathrm{before}}=0.200\mathrm{s}$

The reference point is,$y=0$

As we are given the initial position of the beverage thrown, using kinematic equations we can find the velocity of the beverage when it just reaches the ground. Using the formula for kinetic energy we can find the kinetic energy of the beverage. Using the same procedure, we can find the kinetic energy halfway to the ground.

As we are given the time before it reaches the ground, we find the total time required to reach the ground. We subtract the given time from the required time to get the exact time. From this, we can find the distance traveled during this time, and using the procedure used earlier we can find the kinetic and gravitational potential at that time.

Formulae:

The potential energy at a height, $\mathrm{PE}=\mathrm{mgh}$ (1)

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

The second equation of kinematic motion, $\mathrm{x}={\mathrm{v}}_{\mathrm{i}}\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (3)

The third equation of the kinematic equation,

${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ax}$ (4)

The first equation of kinematic motion, role="math" localid="1661484026511" ${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at}$ (5)

Using the law of conservation of energy

The energy at the initial position = Energy ta the final position,

Thus, using equations (1) and (2), we can give the final kinetic energy as follows:

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2+\mathrm{mgh}=\mathrm{K}.\mathrm{E}{.}_{\mathrm{final}} \\ \mathrm{K}.\mathrm{E}{.}_{\mathrm{final}}=\frac{1}{2}\times 2.5\mathrm{kg}\times {\left(3.0\mathrm{m}/\mathrm{s}\right)}^2+2.5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 4.0\mathrm{m} \\ \mathrm{K}.\mathrm{E}{.}_{\mathrm{final}}=11.25\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2+98\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2 \\ \mathrm{K}.\mathrm{E}{.}_{\mathrm{final}}=109.25\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ \mathrm{K}.\mathrm{E}{.}_{\mathrm{final}}=109.25\mathrm{J} \end{gathered}$$

Hence, the value of the energy is 109.25 J

When the drop height at halfway is h = 2.00 m , the final kinetic energy using the law

of conservation at the hallway is given using equations (1) and (2) as follows:

$$\begin{gathered} \mathrm{K}.\mathrm{E}.=\frac{1}{2}\times 2.5\mathrm{kg}\times {\left(3.0\mathrm{m}/\mathrm{s}\right)}^2+2.50\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 2.0\mathrm{m} \\ =60.3\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =30.3\mathrm{J} \end{gathered}$$

Hence, the kinetic energy at halfway is 60.3 J .

Now, we have to find the kinetic energy time t = 0.200 s before it reaches the ground. Thus, the height at this point can be given using equation (2):

$$\begin{gathered} y=\left(9.35\mathrm{m}/\mathrm{s}\times 0.200\mathrm{s}\right)-\frac{1}{2}{\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)}^2{\left(0.200\mathrm{s}\right)}^2 \\ =1.67\mathrm{m} \end{gathered}$$

Now, the final velocity of the ball at this time is given using equation (5):

$$\begin{gathered} v_f=9.35\mathrm{m}/\mathrm{s}-\left(9.8\mathrm{m}/\mathrm{s}\times 0.200\mathrm{s}\right) \\ =7.40\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the kinetic energy of the ball at this point is given using equation (2):

$$\begin{gathered} \mathrm{KE}=\frac{1}{2}\times 2.5\mathrm{kg}\times {\left(7.40\mathrm{m}/\mathrm{s}\right)}^2 \\ =68.3\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =68.3\mathrm{J} \end{gathered}$$

Hence, the value of the energy is 68.3 J .

Using the height found in part (c) calculations, we can get the gravitational potential using equation (1):

$$\begin{gathered} \mathrm{PE}=2.5\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 1.67\mathrm{m} \\ =41.0\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =41.0\mathrm{J} \end{gathered}$$

Hence, the value of the potential is 41.0 J