Question

Determine the power required for a 1150kg car to climb a 100-m-long uphill road with a slope of 30° (from horizontal) in 12s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance.

Answer 1

Answer:

part (a) 46958.33 W

part (b) 43125 W

part (c) -57500 W

Explanation:

Given,

• Mass of the car = m = 1150 kg
• Angle of inclination of the hill = $\theta\ =\ 30^o$
• Total time taken = t = 12 s

part (a)

Given that the velocity is constant,

Total displacement traveled by the car = s = 100 m

Work done by the car to reach at the top of the hill

$w\ =\ mgsin30^0\times s\\\Rightarrow w\ =\ 1150\times 9.81\times sin30^o\times 100\\\Rightarrow w\ =\ 563500\ J$

Hence, power$P \ =\ \dfrac{w}{t}\\\Rightarrow P\ =\ \dfrac{563500}{12}\\\Rightarrow P\ =\ 46958.33\ W$

part (b)

Given,

• final velocity = $v_f\ =\ 30\ m/s.$
• Initial velocity = $v_i\ =\ 0\ m/s$

Total power is equal to the ratio of the change in kinetic energy and total time taken

$\therefore P\ =\ \dfrac{\Delta K.E}{t}\\\Rightarrow P\ =\ \dfrac{K.E_f\ -\ K.E_i}{t}\\\Rightarrow P\ =\ \dfrac{\dfrac{1}{2}mv_f^2\ -\ 0}{t}\\\Rightarrow P\ =\ \dfrac{mv_f^2}{2t}\\\Rightarrow P\ =\ \dfrac{1150\times 30^2}{2\times 12}\\\Rightarrow P\ =\ 43125\ W$

part (c)

• Initial velocity of the car = $v_i\ =\ 35\ m/s.$
• Final velocity of the car = $v_f\ =\ 5 m/s.$

Total power is equal to the ratio of the change in kinetic energy and total time taken

$\therefore P\ =\ \dfrac{\Delta K.E}{t}\\\Rightarrow P\ =\ \dfrac{K.E_f\ -\ K.E_i}{t}\\\Rightarrow P\ =\ \dfrac{\dfrac{1}{2}mv_f^2\ -\ \dfrac{1}{2}mv_i^2}{t}\\\Rightarrow P\ =\ \dfrac{m(v_f^2\ -\ v_i^2)}{2t}\\\Rightarrow P\ =\ \dfrac{1150\times (5^2\ -\ 35^2)}{2\times 12}\\\Rightarrow P\ =\ -57500\ W$