You are having fun on a water slide. You start from the top of the water slide, which is 2.7 m above the of the water. If you enter the water with a speed of 4.8m/s, how efficient is the water slide in converting your gravitational potential energy into kinetic energy



Answer :

Answer:

To determine the efficiency of the water slide in converting gravitational potential energy (GPE) into kinetic energy (KE), we can use the following steps:

1. **Calculate the initial gravitational potential energy (GPE):**

The gravitational potential energy at the top of the slide is given by:

\[

\text{GPE} = mgh

\]

where:

- \( m \) is the mass of the person (which will cancel out later),

- \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)),

- \( h \) is the height above the water (\( 2.7 \, \text{m} \)).

2. **Calculate the final kinetic energy (KE):**

The kinetic energy when entering the water is given by:

\[

\text{KE} = \frac{1}{2} mv^2

\]

where:

- \( m \) is the mass of the person,

- \( v \) is the velocity when entering the water (\( 4.8 \, \text{m/s} \)).

3. **Calculate the efficiency:**

Efficiency is the ratio of the useful output energy (KE) to the input energy (GPE):

\[

\text{Efficiency} = \frac{\text{KE}}{\text{GPE}} \times 100\%

\]

Let's proceed with these calculations:

1. **GPE Calculation:**

\[

\text{GPE} = mgh = m \times 9.8 \, \text{m/s}^2 \times 2.7 \, \text{m} = 26.46m \, \text{J}

\]

2. **KE Calculation:**

\[

\text{KE} = \frac{1}{2} mv^2 = \frac{1}{2} m \times (4.8 \, \text{m/s})^2 = 11.52m \, \text{J}

\]

3. **Efficiency Calculation:**

\[

\text{Efficiency} = \frac{\text{KE}}{\text{GPE}} \times 100\% = \frac{11.52m \, \text{J}}{26.46m \, \text{J}} \times 100\%

\]

\[

\text{Efficiency} = \frac{11.52}{26.46} \times 100\% \approx 43.5\%

\]

So, the efficiency of the water slide in converting gravitational potential energy into kinetic energy is approximately **43.5%**.

Other Questions