Car Climbing a Gradient at Constant Speed
A car of mass 1200 kg climbs a straight road inclined at an angle θ to the horizontal, where sin θ = 0.10. At a steady speed of 15 m/s, the total resistive force opposing the motion (friction and air resistance combined) is 500 N. The engine delivers power to the driving wheels, and g = 9.81 m/s². The driver then presses the accelerator, so that the engine output immediately becomes 30 kW while the speed is still 15 m/s. The car continues up the same incline, and the resistive force is unchanged at that instant. You should treat the car as a particle and ignore rotational energy in the wheels. The driver wants to know the engine power needed for steady climbing, the initial acceleration when the power is increased, and why the acceleration does not stay constant as the car speeds up.
1. Calculate the engine power required to keep the car climbing at a steady 15 m/s.
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At steady speed the acceleration is zero, so the driving force from the engine must balance the total force opposing motion up the slope. The component of weight down the slope is mg sin θ = 1200 × 9.81 × 0.10 = 1177.2 N. Adding the 500 N resistance gives the required driving force: F = 1177.2 + 500 = 1677.2 N. Power is P = Fv, so P = 1677.2 × 15 = 25,158 W ≈ 25.2 kW. So about 25 kW is needed just to maintain 15 m/s up this incline.
2. Find the initial acceleration when the engine output is raised to 30 kW at 15 m/s.
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At 30 kW and 15 m/s, the driving force is F_d = P / v = 30,000 / 15 = 2000 N. The forces opposing motion up the slope are unchanged at this instant: gravity component 1177.2 N plus resistance 500 N, total 1677.2 N. The net force up the slope is F_net = 2000 − 1677.2 = 322.8 N. By Newton’s second law, a = F_net / m = 322.8 / 1200 = 0.269 m/s². The initial acceleration is therefore about 0.27 m/s² up the slope.
3. Explain why the acceleration decreases as the car’s speed rises even though the engine output remains 30 kW.
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For a fixed power, the driving force falls as speed rises because F_d = P / v. The gravity component and the resistive force are approximately constant in this model, so the net force is F_net = P/v − (mg sin θ + 500). As v increases, P/v decreases, so F_net decreases and hence acceleration decreases. Eventually F_d equals the total opposing force and acceleration becomes zero at a new steady speed. For 30 kW that speed would be v = 30,000 / 1677.2 = 17.9 m/s. This is why a car cannot maintain a constant acceleration while the engine operates at constant power; the acceleration is largest at low speed and falls towards zero as the car approaches its new maximum steady speed on that incline.