Question

ATTACHED BELOW what efficiency is the ball's kinetic energy converted into elastic potential energy

please helpppp urgentttt I'd really appreciate it.


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Answer 1

66.2%

Step-by-step explanation:

The problem at hand involves calculating the efficiency with which a ball's kinetic energy is converted into the elastic potential energy of a spring when the ball collides with and compresses the spring. The efficiency can be determined by comparing the energy stored in the spring (elastic potential energy) with the initial kinetic energy of the ball. We will use the formulas for kinetic energy and efficiency to solve this problem.

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First, we need to calculate the initial kinetic energy (KE) of the ball using the formula for kinetic energy:

`\boxed{\left\begin{array}{ccc}\text{\underline{Kinetic Energy:}}\\\\ K = (1)/(2)mv^2 \\\\\text{Where:}\\\bullet \ K \ \text{is the kinetic energy (J)}\\\bullet \ m \ \text{is the mass of the object (kg)}\\\bullet \ v \ \text{is the velocity of the object (m/s)}\end{array}\right}`

In our case:

• m = 92.4 g = 0.0924 kg
• v₀ = 4.28 m/s

Plugging into our formula:

`\Longrightarrow K=(1)/(2)(0.0924 \text{ kg})(4.28 \text{ m/s})^2\\ \\\\\\\therefore K = 0.846 \ J`

We can now determine the efficiency (η) with which this kinetic energy is converted to elastic potential energy using the formula:

`\eta = \left( \frac{\text{Elastic Potential Energy}}{\text{Kinetic Energy}} \right) * 100\%`

Given that the elastic potential energy stored in the spring at maximum compression is 0.560 J, we can now calculate the efficiency as follows:

`\Longrightarrow \eta = \left( (0.560 \ J)/(0.846 \ J) \right) * 100\%\\\\\\\\\Longrightarrow \eta = 0.662 * 100\%\\\\\\\\\therefore \boxed{ \eta = 66.2\%}`

The ball's kinetic energy is converted to the spring's elastic potential energy with approximately 66.2% efficiency.