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3 years ago

What is the potential energy of a 3 kg rock sitting at the top of a 21 meter high cliff?

Physics

1 answer:

Svetradugi [14.3K]3 years ago

3 0

Answer:

618.03 J

Explanation:

Potential energy can be calculate using the following formula.

PE=m*g*h

where m is the mass, g is the acceleration due to gravity and h is the height.

In this situation, we have a 3 kilogram rock on a 21 meter high cliff. Therefore, the mass is 3 kg and the height is 21 m.

On Earth, the acceleration due to gravity is 9.81 m/s^2.

m=3 kg

h=21 m

g=9.81 m/s^2

Substitute these values into the formula.

PE=3*9.81*21

Multiply all the numbers together

PE=63*9.81

PE=618.03

The potential energy is 618.03 kg m^2/s^2, or 618.03 Joules

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

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Substituting values

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Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

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