Question

A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will it's gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy in m/s?

Answer 1

Answer:

v = 6.3 m/s

Explanation:

• Since no significant air resistance exists, total mechanical energy must be kept constant at any time.
• At the top of the cliff, all the energy is gravitational potential energy, as follows:

       $E_{i} = K_{i} + U_{i} = 0 + U_{i} (1)$

• If we choose the ground level as our zero reference level for the gravitational potential energy, Ui is simply:
• Ui = m*g*h (1)
• At any height, the sum of the kinetic and the gravitational potential energy must be equal to (1).
• We know from the question, that at the point of interest, both types of energies must be equal each other, so we can write the following expression from (1):

       $m*g* h = 2*\frac{1}{2}*m*v^{2} (2)$

• Dividing both sides by m, simplifying, and solving for v, we get:

       $v = \sqrt{g*h} =\sqrt{9.8m/s2*4.0m} = 6.26 m/s (3)$

• v = 6.3 m/s (with two significative figures)