Problem 1: Change in Kinetic Energy of the Rock

We know from the Work-Energy Theorem that the change in the kinetic energy (ΔKE) of the rock is equal to the change in its gravitational potential energy as it falls. This is because all of the potential energy (PE) converts to kinetic energy (KE) if we ignore air resistance.

The change in gravitational potential energy is given by:

$\Delta PE = mgh$

Where:

• $m$ = 1 kg (mass of the rock),
• $g$ = 9.8 m/s² (acceleration due to gravity),
• $h$ = 10 m (height).

Now, substitute these values into the equation:

$\Delta PE = (1\, \text{kg}) (9.8\, \text{m/s}^2) (10\, \text{m}) = 98\, \text{J}$

Since all the potential energy converts to kinetic energy:

$\Delta KE = \Delta PE = 98\, \text{J} \approx 100\, \text{J}$

Thus, the correct answer is 1) About 100 J.

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Problem 2: Energy of a Basketball Thrown Upwards

Part (a): Total Mechanical Energy Expression

The total mechanical energy of the basketball when it reaches its maximum height $h_m$ is the sum of its gravitational potential energy (PE) at the maximum height and its kinetic energy (KE). Since at the maximum height, the velocity is zero, the kinetic energy is zero, and the total mechanical energy is just the potential energy at the maximum height:

$E = mgh_m$

Where:

• $m$ = mass of the ball,
• $g$ = acceleration due to gravity,
• $h_m$ = maximum height the ball reaches.

Thus, the expression for the total mechanical energy is:

$E = m g h_m$

Part (b): Maximum Height Calculation

We can use the principle of conservation of mechanical energy. The initial total mechanical energy when the ball leaves the hand is the sum of its initial kinetic energy and its initial gravitational potential energy at height $h_0$:

$E = \frac{1}{2} m v_0^2 + m g h_0$

Substitute the known values:

• $m = 1 \, \text{kg}$,
• $v_0 = 9.5 \, \text{m/s}$,
• $g = 9.8 \, \text{m/s}^2$,
• $h_0 = 1.9 \, \text{m}$.

$E = \frac{1}{2} (1)(9.5)^2 + (1)(9.8)(1.9)$ $E = \frac{1}{2} (90.25) + (18.62)$ $E = 45.125 + 18.62 = 63.745 \, \text{J}$

Now, at the maximum height $h_m$, all the energy is gravitational potential energy:

$E = m g h_m$

Solving for $h_m$:

$h_m = \frac{E}{mg} = \frac{63.745}{(1)(9.8)} = 6.5 \, \text{m}$

Thus, the maximum height $h_m$ is 6.5 meters.

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Summary of Answers:

1. Problem 1: Change in kinetic energy is approximately 100 J.
2. Problem 2 (a): Total mechanical energy $E = mgh_m$.
3. Problem 2 (b): Maximum height $h_m = 6.5 \, \text{m}$.

Would you like further clarification on any part of the solution?

Here are five related questions for deeper understanding:

1. How does the kinetic energy change if air resistance is considered?
2. What would be the total energy if the ball had a mass of 2 kg?
3. How would the maximum height change if the initial velocity was doubled?
4. Can the work-energy theorem apply if the object doesn’t fall vertically?
5. What happens to the potential energy when the rock hits the ground?

Tip: The conservation of mechanical energy applies only in the absence of external forces like air resistance.