Question

23 a car of mass 1400 kg is travelling on a straight, horizontal road at a constant speed of 25 m s⁻¹. the useful output power from the car engine during this motion is 30 kw. the car then travels up a slope at 2.0° to the horizontal, at the same constant speed. what is the total useful output power of the car engine when travelling up the slope? a 12 kw b 31 kw c 42 kw d 65 kw

Explanation:

Step1: Find the resistive force on horizontal road

Power \( P = Fv \), where \( P = 30\space kW = 30000\space W \), \( v = 25\space m/s \). So resistive force \( F_{res} = \frac{P}{v} = \frac{30000}{25} = 1200\space N \).

Step2: Find the component of weight along the slope

Mass \( m = 1400\space kg \), angle \( \theta = 2.0^\circ \). Component of weight \( F_{g\parallel} = mg\sin\theta \), \( g = 9.8\space m/s^2 \). So \( F_{g\parallel} = 1400\times9.8\times\sin(2.0^\circ) \approx 1400\times9.8\times0.0349 \approx 485\space N \).

Step3: Total force on the slope

Total force \( F_{total} = F_{res} + F_{g\parallel} = 1200 + 485 = 1685\space N \).

Step4: Calculate power on the slope

Power \( P_{slope} = F_{total}v = 1685\times25 = 42125\space W \approx 42\space kW \).

Step5: Total useful output power

Total power \( P_{total} = 30 + 12 = 42\space kW \)? Wait, no, wait. Wait, on horizontal, power is 30 kW (to overcome resistance). On slope, we need to overcome resistance and the component of weight. Wait, no, the 30 kW is the power on horizontal, which is to overcome resistance. When going up slope, the force needed is resistance + component of weight. So power on slope is \( (F_{res} + mg\sin\theta)v \). Let's recalculate:

\( mg\sin\theta = 1400\times9.8\times\sin(2^\circ) \). \( \sin(2^\circ) \approx 0.0348995 \). So \( 1400\times9.8\times0.0348995 \approx 1400\times0.342015 \approx 478.821\space N \). Then total force \( F = 1200 + 478.821 = 1678.821\space N \). Power \( P = Fv = 1678.821\times25 = 41970.525\space W \approx 42\space kW \). Then total power is horizontal power (30 kW) plus the power to overcome gravity? Wait, no, the horizontal power is to overcome resistance. When going up slope, the engine must provide power to overcome resistance AND the component of gravity. So the total power is the power to overcome resistance (which is still 30 kW? Wait, no, the speed is the same, 25 m/s. Wait, maybe I made a mistake. Wait, on horizontal, \( P = F_{res}v = 30\space kW \), so \( F_{res} = 30000/25 = 1200\space N \). On slope, the force required from the engine is \( F_{res} + mg\sin\theta \), because it has to overcome both resistance and the component of weight down the slope. So the power on slope is \( (F_{res} + mg\sin\theta)v \). Let's compute \( mg\sin\theta \):

\( m = 1400\space kg \), \( g = 9.8\space m/s^2 \), \( \theta = 2^\circ \).

\( mg\sin\theta = 1400 \times 9.8 \times \sin(2^\circ) \)

\( \sin(2^\circ) \approx 0.0348995 \)

So \( 1400 \times 9.8 = 13720 \)

\( 13720 \times 0.0348995 \approx 13720 \times 0.0349 \approx 478.828\space N \)

Then total force \( F = 1200 + 478.828 = 1678.828\space N \)

Power \( P = F \times v = 1678.828 \times 25 = 41970.7\space W \approx 42\space kW \)

So total useful output power is the power on horizontal (30 kW) plus the power to overcome gravity? Wait, no, the 30 kW is the power to overcome resistance. When going up slope, the engine has to provide power to overcome resistance (so still 30 kW?) and also power to overcome the component of weight. Wait, no, the speed is constant, so the net force is zero. So the engine force must equal the sum of resistance and the component of weight. So the power is \( F_{engine} \times v = (F_{res} + mg\sin\theta) \times v \). Which is \( F_{res}v + mg\sin\theta \times v \). \( F_{res}v \) is 30 kW, and \( mg\sin\theta \times v \) is the power to overcome gravity. Let's compute \( mg\sin\theta \times v \):

\( 478.828 \times 25 = 11970.7\space W \approx 12\space kW \)

So total power is \( 30 + 12 = 42\space kW \), which matc…

Answer:

C. 42 kW