Question

As a block of mass 42 kilograms drops from the edge of a 40-meter-high cliff it experiences a loss of energy due to air resistance of 81 J. At what speed will the rock hit the ground?

Answer 1

Answer:

The block hits the ground at 27.9 m/s

Explanation:

Gravitational Potential Energy (GPE)

It's the energy stored in an object because of its height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or $9.8 m/s^2$.

When the block is at the edge of the cliff it has potential energy that can be transformed into any other type of energy as it starts falling to the ground.

The GPE of the block of mass m=42 Kg at h=40 m is:

U = 42*9.8*40

U = 16,464 J

The block loses 81 J due to air resistance, thus the energy stored when it hits the ground is 16,464 J - 81 J = 16,383 J.

This energy is stored as kinetic energy, whose formula is:

$\displaystyle K=\frac{1}{2}mv^2$

Solving for v:

$\displaystyle v=\sqrt{\frac{2K}{m}}$

$\displaystyle v=\sqrt{\frac{2*16,383 }{42}}$

$v=\sqrt{780.143}$

v = 27.9 m/s

The block hits the ground at 27.9 m/s