Exam 13: Energy and Power

Correct Answer:

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To estimate the required speed of the bowling ball at point 1 so that it reaches point 2 with a speed of $2 \mathrm {~m} / \mathrm { s }$ , we need to apply the principle of conservation of mechanical energy, assuming there is no friction or air resistance.

The mechanical energy of the ball consists of its kinetic energy (KE) and potential energy (PE). At point 1, the ball will have a certain amount of kinetic energy due to its speed and potential energy due to its height above some reference level. At point 2, the ball will also have kinetic and potential energy.

Let's denote:
- $v_1$ as the speed of the ball at point 1,
- $v_2 = 2 \mathrm {~m} / \mathrm { s }$ as the speed of the ball at point 2,
- $h_1$ as the height of the ball at point 1,
- $h_2$ as the height of the ball at point 2,
- $m$ as the mass of the ball (which will cancel out in the end),
- $g$ as the acceleration due to gravity ( $9.81 \mathrm {~m} / \mathrm { s }^2$ ).

The conservation of mechanical energy states that the total mechanical energy at point 1 is equal to the total mechanical energy at point 2:

$$KE_1 + PE_1 = KE_2 + PE_2$$

$$\frac{1}{2} m v_1^2 + m g h_1 = \frac{1}{2} m v_2^2 + m g h_2$$

Since the mass of the ball $m$ appears in every term, it can be canceled out:

$$\frac{1}{2} v_1^2 + g h_1 = \frac{1}{2} v_2^2 + g h_2$$

Now, we need to solve for $v_1$ :

$$\frac{1}{2} v_1^2 = \frac{1}{2} v_2^2 + g h_2 - g h_1$$

$$v_1^2 = v_2^2 + 2g (h_2 - h_1)$$

$$v_1 = \sqrt{v_2^2 + 2g (h_2 - h_1)}$$

However, the height $h_2$ is not provided in the question. To proceed, we need the value of $h_2 - h_1$ . Assuming $h_2 - h_1$ is given or can be measured, you can plug in the values to find $v_1$ .

For example, if $h_2 - h_1 = 1 \mathrm {~m}$ , then:

$$v_1 = \sqrt{(2 \mathrm {~m} / \mathrm { s })^2 + 2 \cdot 9.81 \mathrm {~m} / \mathrm { s }^2 \cdot 1 \mathrm {~m}}$$

$$v_1 = \sqrt{4 + 19.62}$$

$$v_1 = \sqrt{23.62}$$

$$v_1 \approx 4.86 \mathrm {~m} / \mathrm { s }$$

So, if the height difference is 1 meter, the ball would need to be traveling at approximately $4.86 \mathrm {~m} / \mathrm { s }$ at point 1 to reach point 2 with a speed of $2 \mathrm {~m} / \mathrm { s }$ . If the height difference is different, you would need to adjust the calculation accordingly.