Question

(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.)

Short answer

(a) The average useful power output of the person is 208.33 W.

(b) Time taken to lift the brick to a platform is 141.12 s.

Step-by-step solution

Power and energy consumed

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.

Average useful power output

(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}.

Time required to do work

(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.