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Sagot :
Given
m = 20 kg
h = 100m
Kinetic energy and potential energy after
(i) 1 sec
(ii) 2 sec
Procedure
The motion we are analyzing is a free fall, therefore the acceleration faced by the object is 9.8m/s^2 which is gravity.
Kinetic energy is a form of energy that an object or a particle has by reason of its motion.
Let's first calculate the velocity after 1s
[tex]\begin{gathered} v^{}_f=v^{}_o-gt \\ v_f=-9.8m/s^2\cdot1s \\ v_f=-9.8m/s \end{gathered}[/tex]Now with the velocity, we can calculate the kinetic energy
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ K=\frac{1}{2}20kg\cdot(9.8m/s)^2 \\ K=960.4\text{ J} \end{gathered}[/tex]Now let's calculate the potential energy after 1s. Let's first calculate the displacement after that time
[tex]\begin{gathered} y=v_ot-\frac{1}{2}gt^2 \\ y=-\frac{1}{2}gt^2 \\ y=-\frac{1}{2}\cdot9.8m/s^2\cdot(1s)^2 \\ y=4.9\text{ m} \end{gathered}[/tex]Now let us proceed to calculate the potential energy
[tex]\begin{gathered} U=\text{mgh} \\ U=20kg\cdot9.8m/s^2\cdot(100m-4.9m) \\ U=18639.6\text{ J} \end{gathered}[/tex]For 1s:
K = 960.4 J
U = 18639.6 J
In a very similar way we are going to calculate for the second 2
Let's first calculate the velocity after 2s
[tex]\begin{gathered} v_f=-9.8m/s^2\cdot2s \\ v_f=-19.6m/s \end{gathered}[/tex]Now with the velocity, we can calculate the kinetic energy
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ K=\frac{1}{2}20kg\cdot(19.6m/s)^2 \\ K=3841.6\text{ J} \end{gathered}[/tex]Now let's calculate the potential energy after 2s. Let's first calculate the displacement after that time
[tex]\begin{gathered} y=v_ot-\frac{1}{2}gt^2 \\ y=-\frac{1}{2}gt^2 \\ y=-\frac{1}{2}\cdot9.8m/s^2\cdot(2s)^2 \\ y=19.6\text{ m} \end{gathered}[/tex]Now let us proceed to calculate the potential energy
[tex]\begin{gathered} U=\text{mgh} \\ U=20kg\cdot9.8m/s^2\cdot(100m-19.6m) \\ U=15758.4\text{ J} \end{gathered}[/tex]
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