A rock at the top of a 30 meter cliff has a mass of 25 kg. Calculate the rock’s gravitational potential energy when dropped off the cliff. Assuming that energy is conserved and there is no air friction and gravity is 9.8 m/sec., at the bottom of the cliff the rock’s potential energy is completely converted to kinetic energy. Use the formula for kinetic energy to calculate the rock’s speed at the bottom of the cliff. Show all calculations.

Answers

Answer 1

Answer:
Potential energy =

   (mass) x (gravity) x (height above the reference level) .

Relative to the bottom of the cliff, the potential energy
at the top of the cliff is

   (25kg) x (9.8 m/s²) x (30 meters)

   = (25 x 9.8 x 30) kg-m²/s²

   = 7,350 joules .

Kinetic energy = (1/2) x (mass) x (speed²)

The rock's kinetic energy at the bottom is
the same as its potential energy at the top.

7,350 joules = (1/2) x (25 kg) x (speed²)

Divide each side
by 12.5kg : 7,350 joules/12.5 kg = speed²

   7,350 kg-m²/s² / 12.5kg = speed²

   (7,350 / 12.5) m²/s² = speed²

      588 m²/s² = speed²
Take the square root
of each side:
   Speed = √(588 m²/s²)

   = 24.248... m/s (rounded)

Answer 2

Answer: Lets see:-

We have our formula for potential energy, which we are trying to solve for,

PE = mgh or potential energy = mass * gravity * height

So we know that PE all depends on these.

Height : 30 meters
Mass : 25 kg
Gravity (which is always constant) : 9.8 m/s/s

Now add into formula.

PE = 25*30*9.8
PE = 7350 Joules

Answer: PE = 7350 Joules

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Answers

Hi there!

When finding work (J) given force (N), we use the equation:

W = FdcosФ <-- Only take into account the force in the direction of motion

Plug in the given values:

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$\sf{work~done=811.89~J }$

Explanation:

Here,

Giventhat,

pulls a wagon a distance of 32 meters the handle is above an angle of 20 degrees above horizontal if you pull on it with a force of 27 N

Tofind,

The total work done

Aswe know that,

${\boxed{\sf{W=f.d.cos\theta }}}$

Accordingtothe question,

$W=27(32)cos{20}\n\nw=811.89J$

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Answers

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Answers

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HOPEITHELPEDYOU...

me·chan·i·cal ad·van·tage
noun
the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of a machine.

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Answers

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Answers

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