Question

A 3.50-kN piano is lifted by three workers at a constant speed to an apartment 25.0 m above the street using a pulley system fastened to the roof of the building. Each worker is able to deliver 165 W of power, and the pulley system is 75.0% efficient (so that 25.0% of the mechanical energy is transformed to other forms due to friction in the pulley). Neglecting the mass of the pulley, find the time required to lift the piano from the street to the apartment.

Short answer

The time required to lift the piano from the street to the apartmentis:

$\Delta t=3.39\min$

Step-by-step solution

Defining the instantaneous power and energy formula

The instantaneous power P is defined as the time rate of energy transfer.

$P=\frac{W}{\Delta t}$

Including the types of energy storage and energy transfer that we have discussed gives

$\Delta K+\Delta U+\Delta E_{int}=W_{total}$

where

$P=$Power

$W=$Work done

$\Delta t=$Change in time in second

$\Delta K=$Change in kinetic energy

$\Delta U=$Change in potential energy

$\Delta E_{int}=$Change in internal energy.

Calculating the total work

As the piano is lifted at constant speed up to the apartment, the total work that must be done on it is$\left(\because \Delta E_{int}=0\right)$

$$\begin{gathered} W_{nc}=\Delta K+\Delta U_g \\ {W}_{nc}=0+mg\left(y_f-y_i\right) \\ {W}_{nc}=\left(3.50\times {10}^{3}N\right)\left(25.0m\right) \\ =8.75\times {10}^{4}J \end{gathered}$$

The three workmen (using a pulley system with an efficiency of 0.750) do work on the piano at a rate of

$$\begin{gathered} P_{net}=0.750\left(3P_{\sin glewor\ker }\right) \\ =0.750\left[3\left(165W\right)\right] \\ =371W=371J/s \end{gathered}$$

So, the time required to do the necessary work on the piano is

$$\begin{gathered} \Delta t=\frac{W_{nc}}{P_{net}} \\ =\frac{8.75\times {10}^{4}J}{371J/s}=236s=236s.\frac{1\min }{60s}=3.39\min \end{gathered}$$