Physics, 08.12.2020 02:40 nikiloum

A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fallen 10 m?

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Answer from: mayasofia

The kinetic and potential energy when the rock that has fallen 10 m will be 981 J and 981 J, respectively.

Given information:

A 10 kg rock falls from a 20 m cliff.

At the height of 20 m, the kinetic energy of the rock will be zero because it is at rest. Now, while falling, its potential energy will decrease and kinetic energy will increase.

According to energy conservation, the loss in potential energy from 20 m to 10 m will be converted into kinetic energy.

So, the loss in potential energy or gain in kinetic energy from 20 m to 10 m will be,

$\Delta PE=mgh\\\Delta PE=10\times 9.81\times (20-10)\\\Delta PE=981\rm\; J$

So, the kinetic energy of the rock at 10 m height will be 981 J.

Now, the potential energy of the rock at 10 m height will be,

$\Delta PE=mgh\\\Delta PE=10\times 9.81\times 10\\\Delta PE=981\rm\; J$

So, the potential energy of the rock at 10 m height will be 981 J.

Therefore, the kinetic and potential energy when the rock that has fallen 10 m will be 981 J and 981 J, respectively.

For more details, refer to the link:

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Answer from: rainbow6e

Hello!

First, lets calculate potential energy at the 10 meters, for that, lets applicate formula:

$\boxed{E_p = mgh}$

$\textbf{Being:}$

$\sqrt{}$ $E_p = Potential\ Energy = \ ?$

$\sqrt{}$ $m = Mass = 10 \ kg$

$\sqrt{}$ $g = Gravity = 9,81\ m/s^{2}$

$\sqrt{}$ $h = Height = 10 \ m$

$\textbf{So let's replace and resolve it}$

$E_p = 10 \ kg * 9,81 \ m/s^{2} * 10 \ m$

$E_p = 981 \ J$

$\text{The potential energy is of \textbf{981 J}}$

Now, lets calculate the kinetic energy, for this, we have the equation:

$\boxed{k = 0,5 * m * v^{2}}$

$\textbf{Being:}$

$\sqrt{}$ $k = Kinetic\ Energy = \ ?$

$\sqrt{}$ $m = Mass = 10 \ kg$

$\sqrt{}$ $v = Velocity = \ ?$

How you see, we have to calculate the velocity in that instant, for that lets applicate formula:

$\boxed{V^{2} = V_o^{2} +2g (\triangle h) }$

$\textbf{So let's replace and resolve it}$

$V^{2} = (0\ m/s)^{2} + 2(9,81 \ m/s ^ {2} ) (20 \ m - 10 \ m)$

$V = \sqrt{196,2 \ m^{2}/s^{2} }$

$V = 14,007 \ m/s$

$\text{Now we have velocity, so let's \textbf{ replace} according kinetic formula:}$

$k = 0,5 * 10 \ kg * (14,007\ m/s)^{ 2 }$

$k = 5\ kg * 196,19\ m^{2}/s^{2}$

$k = 980,98 \ J$

$\text{The kinetic energy is \textbf{980,98 Joules}}$

Good Luck!!

Answer from: Quest

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

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A 10 kg rock falls from a 20 m cliff. What is the kinetic and potential energy when the rock has fa...