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