Problem — Ranking powers (four lifts)

Ranking: powers. Four jobs carried out in different times: (A) lifting $100\;\mathrm{kg}$ by $1\;\mathrm{m}$ in $10\;\mathrm{s}$; (B) lifting $50\;\mathrm{kg}$ by $5\;\mathrm{m}$ in $20\;\mathrm{s}$; (C) lifting $10\;\mathrm{kg}$ by $10\;\mathrm{m}$ in $5\;\mathrm{s}$; (D) lifting $1\;\mathrm{kg}$ by $100\;\mathrm{m}$ in $200\;\mathrm{s}$. Rank the powers from smallest to largest.

Solution

Power is work divided by time. The work of lifting is $W = mgh$, so

$P = \frac{W}{t} = \frac{mgh}{t} \qquad (g = 9{,}81\;\mathrm{m/s^2})$

$P_A = \frac{100\cdot 9{,}81\cdot 1}{10} = 98{,}1\;\mathrm{W}$

$P_B = \frac{50\cdot 9{,}81\cdot 5}{20} = \frac{2452{,}5}{20} = 122{,}6\;\mathrm{W}$

$P_C = \frac{10\cdot 9{,}81\cdot 10}{5} = \frac{981}{5} = 196{,}2\;\mathrm{W}$

$P_D = \frac{1\cdot 9{,}81\cdot 100}{200} = \frac{981}{200} = 4{,}9\;\mathrm{W}$

Ordering $4{,}9 < 98{,}1 < 122{,}6 < 196{,}2$:

$\ev{D < A < B < C}$

Physical reading. All cases have a comparable total work (about $980\;\mathrm{J}$, except B which does $2450$): it is time that decides. C wins because it does a large amount of work in the shortest time ($5\;\mathrm{s}$); D loses because it spreads the work over $200\;\mathrm{s}$. Power measures the rate at which energy is transferred, not the total amount.

Collegamenti

Argomenti: Work and energy Concetti: Power Competenze: Ranking reasoning