Problem — Height gained by a lift

Lift motor. A $1500\;\mathrm{W}$ motor delivers work to a lift for $30\;\mathrm{s}$. Knowing it carries $400\;\mathrm{kg}$, calculate the height gained.

Solution

Data: $P = 1500\;\mathrm{W}$, $t = 30\;\mathrm{s}$, $m = 400\;\mathrm{kg}$.

Work delivered by the motor (power times time):

$W = P\,t = 1500\cdot 30 = 45\,000\;\mathrm{J}$

Height. Assuming all the work becomes gravitational potential energy ($W = mgh$):

$h = \frac{W}{mg} = \frac{45\,000}{400\cdot 9{,}81} = \frac{45\,000}{3924} \approx 11{,}5\;\mathrm{m}$

$\ev{h \approx 11{,}5\;\mathrm{m} \ (\sim 4\ \text{storeys})}$

Physical reading. In $30\;\mathrm{s}$ a modest motor lifts $400\;\mathrm{kg}$ by about 4 storeys. The calculation is an ideal limit: in reality, guide friction and resistances reduce the actual height, so $11{,}5\;\mathrm{m}$ is the theoretical maximum.

Collegamenti

Argomenti: Work and energy Concetti: Power · Gravitational potential energy Oggetti: Lift