GWCB001 Question 3

Answer: 9878400

1. Calculation of Initial Potential Energy

The potential energy $U$U due to gravity when Person A is at the top of the slide can be expressed as follows using the mass $m_0$m0​, gravitational acceleration $g$g, and height $h$h.

$U=m_0gh$U=m0​gh

Substitute the given values into the equation.

$U=72\times9.8\times20=14112\ \mathrm{J}$U=72×9.8×20=14112 J

2. Calculating the Kinetic Energy Just Before Reaching the Bottom

According to the problem statement, $\mu\%$μ% of the total energy is lost as frictional heat while sliding down the slide. Therefore, the kinetic energy $K$K remaining just before reaching the bottom is $70\%$70% of the initial potential energy, or $(1-\mu)$(1−μ).

$K=U\times(1-\mu)$K=U×(1−μ)

$K=14112\times(1-0.30)=14112\times0.70=9878.4\ \mathrm{J}$K=14112×(1−0.30)=14112×0.70=9878.4 J

*Note: Since the loss ratio $\mu$μ is given directly as a percentage of the total energy, we will not use the 30° angle of inclination in this calculation.

3. Convert to the Specified Answer Format

Multiply the calculated kinetic energy $K$K by 1000.

$Desired value = K \times 1000 = 9878.4 \times 1000 = 9878400$Desiredvalue=K×1000=9878.4×1000=9878400

Bonus

I wonder what would happen if you used just this energy lost as frictional heat to heat 1 mL of water to 300 K (Kelvin)?