A mass of 15 kg falls freely from rest and reaches a velocity of 23 m/s just as it reaches the ground. Calculate: 4.3.1 The potential energy when the mass is at rest. 4.3.2 The height through which it falls.

QUESTION 4.3: ENERGY

A mass of $15 \text{ kg}$ falls freely from rest and reaches a velocity of $23 \text{ m/s}$ just as it reaches the ground. Given:

• Mass $m = 15 \text{ kg}$
• Initial velocity $v_i = 0 \text{ m/s}$ (falls from rest)
• Final velocity $v_f = 23 \text{ m/s}$
• Acceleration due to gravity $g = 9.81 m/s^2$

4.3.1 The potential energy when the mass is at rest.

Step 1: Apply the principle of conservation of energy. When the mass is at rest at its maximum height, it possesses only potential energy. Just before it hits the ground, all this potential energy has been converted into kinetic energy (assuming no air resistance). Therefore, the initial potential energy ($PE_i$) is equal to the final kinetic energy ($KE_f$). $PE_i = KE_f = \frac{1}{2}mv_f^2$ Substitute the given values: $PE_i = \frac{1}{2} \times 15 kg \times (23 m/s)^2$ $PE_i = \frac{1}{2} \times 15 kg \times 529 m^2/s^2$ $PE_i = 7.5 \times 529 \text{ J}$ $PE_i = 3967.5 \text{ J}$ Rounding to three significant figures: $\boxed{3970 J}$

4.3.2 The height through which it falls.

Step 1: Use the kinematic equation relating initial velocity, final velocity, acceleration, and displacement. We know $v_f^2 = v_i^2 + 2gh$, where $h$ is the height and $g$ is the acceleration due to gravity. Since the mass falls from rest, $v_i = 0 \text{ m/s}$. $v_f^2 = 0^2 + 2gh$ $v_f^2 = 2gh$ Rearrange to solve for $h$: $h = \frac{v_f^2}{2g}$ Substitute the given values: $h = \frac{(23 m/s)^2}{2 \times 9.81 m/s^2}$ $h = \frac{529 m^2/s^2}{19.62 m/s^2}$ $h \approx 26.962 \text{ m}$ Rounding to three significant figures: $\boxed{27.0 m}$