Question

prephysics efficiency practice

1. you put 750 j of energy into a system by raising a 15 kg block.

a) how far can you raise the block with 100% efficiency?
b) how far can you raise the block with 85% efficiency?

1. you put 7500 j of energy into a system by raising a 15 kg block by 12 m.

a) how much energy should have been required?
b) if you actually used 7500 j, what was the efficiency of the system?

1. you have a block with a mass of 5 kg. the block is currently moving at 25 m/s.

a) how much ke does the block have?
b) if you would like to convert the energy to gravitational energy, how high can it be raised?
c) with 75% efficiency, how far can the block be raised?

Explanation:

Step1: Define useful energy (100% efficiency)

At 100% efficiency, input energy equals gravitational potential energy: $E_{in} = mgh$. Rearrange for height $h$:
$h = \frac{E_{in}}{mg}$

Step2: Calculate height for 1a

Substitute $E_{in}=750\ \text{J}$, $m=15\ \text{kg}$, $g=9.8\ \text{m/s}^2$:
$h = \frac{750}{15 \times 9.8} = \frac{750}{147} \approx 5.10\ \text{m}$

Step3: Define useful energy (85% efficiency)

At 85% efficiency, useful energy $E_{out} = 0.85 \times E_{in}$. Use $E_{out}=mgh$, rearrange for $h$:
$h = \frac{0.85 \times E_{in}}{mg}$

Step4: Calculate height for 1b

Substitute values:
$h = \frac{0.85 \times 750}{15 \times 9.8} = \frac{637.5}{147} \approx 4.34\ \text{m}$

Step5: Calculate required energy for 2a

Required energy is gravitational potential energy: $E_{req} = mgh$
Substitute $m=15\ \text{kg}$, $h=12\ \text{m}$, $g=9.8\ \text{m/s}^2$:
$E_{req} = 15 \times 9.8 \times 12 = 1764\ \text{J}$

Step6: Calculate efficiency for 2b

Efficiency $\eta = \frac{E_{req}}{E_{used}} \times 100\%$
Substitute $E_{req}=1764\ \text{J}$, $E_{used}=7500\ \text{J}$:
$\eta = \frac{1764}{7500} \times 100\% = 23.52\%$

Step7: Confirm KE for 3a

Kinetic energy formula: $KE = \frac{1}{2}mv^2$
Substitute $m=5\ \text{kg}$, $v=25\ \text{m/s}$:
$KE = \frac{1}{2} \times 5 \times 25^2 = \frac{1}{2} \times 5 \times 625 = 1562.5\ \text{J}$

Step8: Calculate height for 3b

Set $KE = mgh$, rearrange for $h$:
$h = \frac{KE}{mg}$
Substitute values:
$h = \frac{1562.5}{5 \times 9.8} = \frac{1562.5}{49} \approx 31.89\ \text{m}$

Step9: Calculate height for 3c

At 75% efficiency, useful energy $E_{out}=0.75 \times KE$. Use $E_{out}=mgh$, rearrange for $h$:
$h = \frac{0.75 \times KE}{mg}$
Substitute values:
$h = \frac{0.75 \times 1562.5}{5 \times 9.8} = \frac{1171.875}{49} \approx 23.92\ \text{m}$

Answer:

1a) $\approx 5.10\ \text{m}$
1b) $\approx 4.34\ \text{m}$
2a) $1764\ \text{J}$
2b) $23.52\%$
3a) $1562.5\ \text{J}$
3b) $\approx 31.89\ \text{m}$
3c) $\approx 23.92\ \text{m}$