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A stone of mass 2.0kg is dropped from a height of 4.0m. Neglecting air resistance, calculate the kinetic energy of thr stone just before it reaches the ground

User Colmulhall
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4 votes

Answer:

78.4 Joules

Step-by-step explanation:

Potential Energy

Lets start with the stone's potential energy (PE) at a height of 4.0 m above Earth's surface. PE due to Earth's gravity, g,

PE = mgh, where m is the mass, g is the acceleration due to Earth's gravity, and h is the height. g is 9.8 m/s^2.

PE = (2.0 kg)*(9.8 m/s^2)*(4.0 m)

PE = 78.4 kg*m^2/s^2

Note that 1 Joule(J) is equal to 1 kg*m^2/s^2, so we can write

PE = 78.4 J

We might be tempted to stop here, making the assumption that all of this potential energy will be converted into kinetic energy at the time the stone hits Earth. But let's see if we can confirm this assumption.

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Kinetic Energy

The ball will accelerate towards Earth once it is released. Assuming no air friction, we can calculate it's velocity after it falls 4 meters (just touching Earth's surface).

The distance a falling object travels can be calculated using the formula:

d=0.5*g*t^2 where d is the distance, g is the acceleration of gravity (-9.8 m/s^2), and t is time (in seconds here).

Lets rearrange to find the time required to travel 4 meters towards Earth:

d=0.5*g*t^2

2d = g*t^2

t^2 = 2d/g

t^2 = 2*(4 m)/(-9.8 m/s^2) [The minus sign simply indicates the direction of the object (towards Earth)]

t^2 = (8 m)/(-9.8 m/s^2)

t^2 = 0.82 s

t = 0.904 seconds

The stone is falling for 9.04 seconds before hitting Earth. Let's now calculate the stone's speed at that time:

v = a*t

Velocity = (acceleration)*(time)

v = (9.8 m/s^2)*(0.904 seconds)

v = 8.85 m/s

The stone's kinetic energy (KE) at this point is given by

KE = (1/2)mv^2

KE = (1/2)*(2.0 kg)(8.85 m/s)^2

KE = 78.4 kg*m/s^2 or Joules

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This is the same amount of potential energy we calculated at the start. Either calculation would have sufficed, but doing both is more entertaining.

User Fastobject
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