A 500-kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s?

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

Mathematically,

\(P = \frac{W}{T}\)

Here, W is the net work done and T is the time taken.

The work done by kinetic energy is,

\({W_k} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\)

Here, \(m\)is the mass of the dragster \(\left( {m = 500{\rm{ kg}}} \right)\), \({v_f}\) is the final velocity of the dragster \(\left( {{v_f} = 110{\rm{ m}}/{\rm{s}}} \right)\), and \({v_i}\) is the initial velocity of the dragster \(\left( {{v_i} = 0} \right)\).

Putting all known values,

\(\begin{aligned}{}{W_k} &= \frac{1}{2} \times \left( {500{\rm{ kg}}} \right) \times {\left( {110{\rm{ m}}/{\rm{s}}} \right)^2} - \frac{1}{2} \times \left( {500{\rm{ kg}}} \right) \times {\left( 0 \right)^2}\\ &= 3.025 \times {10^6}{\rm{ J}}\end{aligned}\)

The work done by the frictional force is,

\({W_f} = Fd\)

Here, F is the average frictional force \(\left( {1200{\rm{ N}}} \right)\) and d is the displacement \(\left( {d = 400{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned}{}{W_f} &= \left( {1200{\rm{ N}}} \right) \times \left( {400{\rm{ m}}} \right)\\ &= 4.8 \times {10^5}{\rm{ J}}\end{aligned}\)

The total work done is,

\(\begin{aligned}{}W &= {W_k} + {W_f}\\ &= \left( {3.025 \times {{10}^6}{\rm{ J}}} \right) + \left( {4.8 \times {{10}^5}{\rm{ J}}} \right)\\ &= 3.505 \times {10^6}{\rm{ J}} \end{aligned}\)

The power is defined as,

\(P = \frac{W}{T}\)

Here, W is the net work done and T is the time taken \(\left( {T = 7.3{\rm{ s}}} \right)\).

Putting all known values,

\(\begin{aligned}{}P &= \frac{{3.505 \times {{10}^6}{\rm{ J}}}}{{7.3{\rm{ s}}}}\\& = 4.8 \times {10^{\rm{5}}}{\rm{ W}}\end{aligned}\)

Converting power into horsepower\(\left( {{\rm{Hp}}} \right)\),

\(\begin{aligned}{}P &= 4.8 \times {10^5}{\rm{ W}}\\ &= \left( {4.8 \times {{10}^5}{\rm{ W}}} \right) \times \left( {\frac{{1{\rm{ Hp}}}}{{746{\rm{ W}}}}} \right)\\ &= 643.43{\rm{ Hp}}\end{aligned}\)

Therefore, the required average power output is \(4.8 \times {10^5}{\rm{ W}}\) or \(643.43{\rm{ Hp}}\).