Question

c) if the energy is used to launch a 5 kg block into the air, how high can the block be flung? 3) a block has a mass of 2.4 kg and is traveling at 20 m/s. after an energy conversion, it stopped 25 meters higher than where it started. is this possible in the real world? justify your answer using a calculation of efficiency.

Explanation:

First, addressing part c):

Step1: Define energy equivalence

Kinetic energy becomes gravitational potential energy: $KE = PE$

Step2: Write energy formulas

$\frac{1}{2}mv^2 = mgh$

Step3: Cancel mass, solve for h

$h = \frac{v^2}{2g}$ (Note: To solve fully, we need the initial energy/velocity of the block. Assuming we use the velocity from part 3, $v=20\ \text{m/s}$, $g=9.8\ \text{m/s}^2$)
$\displaystyle h = \frac{(20)^2}{2\times9.8}$

Step4: Calculate final height

$\displaystyle h = \frac{400}{19.6} \approx 20.41\ \text{m}$

Step1: Calculate initial kinetic energy

$KE_{initial} = \frac{1}{2}mv^2$
$\displaystyle KE_{initial} = \frac{1}{2}\times2.4\times(20)^2 = 480\ \text{J}$

Step2: Calculate final potential energy

$PE_{final} = mgh$
$\displaystyle PE_{final} = 2.4\times9.8\times25 = 588\ \text{J}$

Step3: Calculate energy efficiency

$\text{Efficiency} = \frac{\text{Useful Output Energy}}{\text{Input Energy}} \times 100\%$
$\displaystyle \text{Efficiency} = \frac{588}{480} \times 100\% = 122.5\%$

Answer:

$\approx 20.41$ meters

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Now addressing question 3):