Analyzing Free Fall Kinetic Energy Comparison

This question asks us to compare the kinetic energy (KE) of a falling body at two different times, 3 seconds and 4 seconds, after it starts falling from rest.

• Key Concepts: Free fall means the only force acting is gravity, causing constant downward acceleration (g). Kinetic energy depends on mass and the square of velocity ($KE = \frac{1}{2}mv^2$).
• Initial Conditions: Mass ($m$) = 1 Kg, Initial velocity ($u$) = 0 m/s (starts from rest).

Calculating Velocities During Free Fall

We use the first equation of motion to find the velocity ($v$) at time ($t$): $v = u + at$. Here, $a = g$ (acceleration due to gravity).

• Velocity at t = 3 seconds:

  $v_3 = 0 + g \times 3 = 3g$

• Velocity at t = 4 seconds:

  $v_4 = 0 + g \times 4 = 4g$

Since $g$ is positive, $v_4 > v_3$. The velocity at 4 seconds is greater than the velocity at 3 seconds.

Comparing Kinetic Energies

Kinetic energy is calculated as $KE = \frac{1}{2}mv^2$. Since the mass ($m$) is constant (1 Kg) and velocity ($v$) increases with time in free fall, the kinetic energy also increases with time.

• Kinetic Energy at t = 3 seconds:

  $KE_3 = \frac{1}{2} \times 1 \times (v_3)^2 = \frac{1}{2} \times (3g)^2 = \frac{9g^2}{2}$

• Kinetic Energy at t = 4 seconds:

  $KE_4 = \frac{1}{2} \times 1 \times (v_4)^2 = \frac{1}{2} \times (4g)^2 = \frac{16g^2}{2}$

Comparing $KE_4$ and $KE_3$: $KE_4 = \frac{16g^2}{2}$ and $KE_3 = \frac{9g^2}{2}$. Clearly, $KE_4 > KE_3$.

Conclusion

The kinetic energy at 4 seconds is greater than the kinetic energy at 3 seconds because the velocity increases during free fall.