(a) What is the average useful power output of a person who does \begin{aligned}6.00 \times {10^6}{\rm{ J}} \end{aligned}of useful work in 8.00 h?

(b) Working at this rate, how long will it take this person to lift 2000 kgof bricks 1.50 mto a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)

The power or rating of a device is defined as the energy consumed by the appliances in a given time.

Power is defined as,

\[P = \frac{E}{T}\]

Here, E is the energy consumed and T is the time consumed.

(a)

When a person does \begin{aligned}6.00 \times {10^6}{\rm{ J}} \end{aligned} of work in 8.00 h, then the average useful output power of the person is,

\begin{aligned}P &= \frac{{6.00 \times {{10}^6}{\rm{ J}}}}{{8.00{\rm{ h}}}}\\ &= \frac{{6.00 \times {{10}^6}{\rm{ J}}}}{{\left( {8.00{\rm{ h}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right)}}\\ &= 208.33{\rm{ W}} \end{aligned}

Therefore, the required average useful power output of the person is \begin{aligned}208.33{\rm{ W}} \end{aligned}.

(b)

The energy required to lift 2000 kg of bricks to a 1.5 m high platform is,

\begin{aligned}{E_P} = mgh \end{aligned}

Here, m is the mass of the brick, g is the acceleration due to gravity \begin{aligned}\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \end{aligned}, and h is the height of the platform.

Putting all known values,

\begin{aligned}{E_P} &= \left( {2000{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {1.5{\rm{ m}}} \right)\\ &= 29400{\rm{ J}} \end{aligned}

The time required to lift the brick to the platform at the of 208.33 W is,

\begin{aligned}T = \frac{{{E_P}}}{P} \end{aligned}

Putting all known values,

\begin{aligned}T &= \frac{{29400{\rm{ J}}}}{{208.33{\rm{ W}}}}\\ &= 141.12{\rm{ s}} \end{aligned}

Therefore, the time taken to lift the brick to a platform is 141.12 s.