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A stone of mass 1.2kg is thrown vertically downwards from a building with an initial speed of 2.0 m/s. The stone loses 96J of gravitational potential energy as it hits the ground. Acceleration due to gravity is 10 m/s. Energy lost due to work done against air resistance is 20J. What is the velocity of the stone just before it hits the ground?

Hello Isabel,

As I explain this, I will leave some of the math for you to do. At the moment of leaving the top of the building the stone possesses both initial kinetic energy, KEi, due to the throw, and gravitational potential energy, GPEi, due to being at the top of the building. The initial kinetic energy is given by

KEi = (1/2)*m*v^2 = (1/2)*1.2 kg*(2.0 m/s)^2

When you multiply that out, the units of the result will be Joules. The problem says that the stone looses 96 J of GPE on the way down. We can consider ground level to be the reference level, the point at which GPE is zero, so at roof level

GPEi = 96 J.

So the total initial energy,

Eti = KEi + GPEi.

On the way down, the stone looses 20 J of energy due to air resistance. Therefore, as it finally arrives at ground level, the stone's final total energy, Etf, is

Etf = KEi + GPEi - 20 J

The stone has no gravitational potential energy as it reaches the ground, so the stone's total energy, Etf, is entirely kinetic energy, KEf. Therefore KEf and Etf are equal.

So we can say that

KEi + GPEi - 20 J = (1/2)*m*v^2

Plug in the values for KEi, GPEi, and mass, then solve for velocity.

I hope this helps,

Steve

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I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.**Education/Credentials**

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