Given:
The mass of the bean is
[tex]\begin{gathered} m=2.0\text{ g} \\ m=2.0\times10^{-3}\text{ kg} \end{gathered}[/tex]The distance during which it traveled is
[tex]\begin{gathered} h=1.0\text{ cm} \\ h=0.010\text{ m} \end{gathered}[/tex]Required: (a) Potential energy gained
(b) the speed of the bean when lands back in the palm.
Explanation:
we have to apply conservation of energy to solve the problem.
when the bean is in our hand it has zero potential energy and after reaching its highest point has only potential energy.
look at the free body diagram
The potential energy is given as
[tex]P.E=mgh[/tex]here,
[tex]h[/tex]is the height and
[tex]g\text{ }[/tex]is the acceleration due to gravity which is equal to
[tex]9.8\text{ m/s}^2[/tex]Plugging all the values in the above relation, we get
[tex]\begin{gathered} P.E=mgh \\ P.E=2.0\times10^{-3}\text{ kg}\times9.8\text{ m/s}^2\times0.010\text{ m} \\ P.E=0.196\times10^{-3}\text{ J} \\ P.E=1.96\times10^{-4}\text{ J} \end{gathered}[/tex](a) Potential energy is
[tex]1.96\times10^{-4}\text{ J}[/tex](b)
By energy conservation,
when a bean comes down in the hand, total potential energy is converted into kinetic energy.
[tex]K.E=P.E[/tex]Kinetic energy is given as
[tex]K.E=\frac{1}{2}mv^2[/tex]put this into the above relation, we get
[tex]\begin{gathered} K.E=P.E \\ \frac{1}{2}mv^2=P.E \\ \end{gathered}[/tex][tex]v=\sqrt[2]{2\times\frac{P.E}{m}}[/tex]Plugging all the values in the above relation, we get
[tex]\begin{gathered} v=\sqrt[2]{2\times1.96\times\frac{10^{-\frac{4}{}}\text{ J}}{2\times10^{-3}\text{ kg}}} \\ \\ v=\sqrt[2]{1.96\times10^{-1}} \\ v=0.44\text{ m/s} \end{gathered}[/tex]Thus, the speed of the bean as lands back in the palm is
[tex]0.44\text{ m/s}[/tex]
