Calculate the power output needed for a 950-kg car to climb a 2.00° slope at a constant 30.0 m/s while encountering wind resistance and friction totaling 600 N. Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy.

Power: Power is a scalar quantity which is defined how fast the energy is being used.

The power is defined as,

\(P = \frac{W}{t}\) !(.1)

Here, W is the work done, and t is the time.

The work done is,

\(W = Fs\) (1.2)

Here, F is the total force, and s is the displacement.

From equation (1.1) and (1.2),

\(\begin{aligned}P &= \frac{{Fs}}{t}\\ &= F\frac{s}{t}\\ &= Fv\end{aligned}\) (1.3)

Since, the velocity is,

\(v = \frac{s}{t}\)

Total force on the body:

Here, m is the mass of the car\(\left( {m = 950{\rm{ kg}}} \right)\), g is the acceleration due to gravity\(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\),\(\theta \) is the angle of inclination\(\left( {\theta = {{2.00}^ \circ }} \right)\), f is the friction and wind resistance\(\left( {f = 600{\rm{ N}}} \right)\), F is the total force.

The force equation of the body is,

\(F = mg\sin \theta + f\)

Putting all known values,

\(\begin{aligned}F& = \left( {950{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \sin \left( {{{2.00}^ \circ }} \right) + \left( {600{\rm{ N}}} \right)\\ &= 924.91{\rm{ N}}\end{aligned}\)

If the body is moving with\(v = 30.0{\rm{ m}}/{\rm{s}}\). The power output needed is calculated using equation (1.3).

Putting all known values,

\(\begin{aligned}P& = \left( {924.91{\rm{ N}}} \right) \times \left( {30.0{\rm{ m}}/{\rm{s}}} \right)\\& = 27.7 \times {10^3}{\rm{ N}}\\ &= 27.7{\rm{ kN}}\end{aligned}\)

Therefore, the required power output is \(27.7{\rm{ kW}}\).