Question

1. a 17 kg block is currently 9.2 meters off the ground.

a) how much gravitational energy does it have?
b) how much energy can be converted with an 87% efficiency?
c) how far can you compress a spring with a k=175 n/m with this energy?

1. a 37 kg block is 25 meters off the ground. you use the energy of the block to compress a spring with a k=755 n/m. how far will the block compress the spring with 95% efficiency?
2. a 25 n/m spring is compressed by 2.3 meters. how fast can it accelerate a 2.1 kg block with 89% efficiency?
3. a biker has a mass of 150 kg with their bike and is biking at 25 m/s. they stop pedaling to roll up a hill, rolling 28 meters vertically up the hill. what was the efficiency of the bike?

Explanation:

Problem 4

Step1: Calculate gravitational potential energy

Use formula $U_g = mgh$
$U_g = 17\ \text{kg} \times 9.8\ \text{m/s}^2 \times 9.2\ \text{m} = 1532.72\ \text{J}$

Step2: Find usable energy (87% efficiency)

Multiply $U_g$ by efficiency
$E_{\text{usable}} = 1532.72\ \text{J} \times 0.87 = 1333.47\ \text{J}$

Step3: Solve for spring compression

Set usable energy equal to spring potential energy $E_{\text{usable}} = \frac{1}{2}kx^2$, solve for $x$
$x = \sqrt{\frac{2E_{\text{usable}}}{k}} = \sqrt{\frac{2 \times 1333.47\ \text{J}}{175\ \text{N/m}}} \approx 3.91\ \text{m}$

Step1: Calculate gravitational potential energy

Use $U_g = mgh$
$U_g = 37\ \text{kg} \times 9.8\ \text{m/s}^2 \times 25\ \text{m} = 9065\ \text{J}$

Step2: Find usable energy (95% efficiency)

$E_{\text{usable}} = 9065\ \text{J} \times 0.95 = 8611.75\ \text{J}$

Step3: Solve for spring compression

Set $E_{\text{usable}} = \frac{1}{2}kx^2$, solve for $x$
$x = \sqrt{\frac{2 \times 8611.75\ \text{J}}{755\ \text{N/m}}} \approx 4.76\ \text{m}$

Step1: Calculate spring potential energy

Use $U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2} \times 25\ \text{N/m} \times (2.3\ \text{m})^2 = 66.125\ \text{J}$

Step2: Find usable energy (89% efficiency)

$E_{\text{usable}} = 66.125\ \text{J} \times 0.89 = 58.851\ \text{J}$

Step3: Solve for block speed

Set $E_{\text{usable}} = \frac{1}{2}mv^2$, solve for $v$
$v = \sqrt{\frac{2 \times 58.851\ \text{J}}{2.1\ \text{kg}}} \approx 7.47\ \text{m/s}$

Answer:

a) $1532.72\ \text{J}$
b) $1333.47\ \text{J}$
c) $\approx 3.91\ \text{m}$

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Problem 5