3. A 10 kg rock rests on top of a cliff 10 meters above the earth's surface.If the rock falls and all of its energy is conserved what is its kinetic energy the second just before it hits the ground? (Show work) 4. 0J. B . 98J C. 480J D. 980 J

## Step 1<br />The problem involves the principle of conservation of energy, which states that the total energy in an isolated system remains constant. In this case, the rock's potential energy is converted into kinetic energy as it falls.<br /><br />## Step 2<br />The potential energy (PE) of the rock when it is at rest on the cliff can be calculated using the formula:<br />### \( PE = m \cdot g \cdot h \)<br />where \( m \) is the mass of the rock, \( g \) is the acceleration due to gravity, and \( h \) is the height of the cliff.<br /><br />## Step 3<br />Substituting the given values into the formula, we get:<br />### \( PE = 10 \, kg \cdot 9.8 \, m/s^2 \cdot 10 \, m = 980 \, J \)<br /><br />## Step 4<br />As the rock falls, all of its potential energy is converted into kinetic energy. Therefore, the kinetic energy of the rock just before it hits the ground is also 980 J.