The mechanical system of a 2.85 x 103kg truck is running at 24.5% efficiency when it is driving the truck ata speed of 42.0 km/h up a uniform incline. If the driving force exerted by the wheels on the road is 2.40 x 103N, what power must the truck's motor generate?a. 158 kW c. 114 kWb. 134 kW d. 99.5 kW
Question
The mechanical system of a 2.85 x 103kg truck is running at 24.5% efficiency when it is driving the truck ata speed of 42.0 km/h up a uniform incline. If the driving force exerted by the wheels on the road is 2.40 x 103N, what power must the truck's motor generate?a. 158 kW c. 114 kWb. 134 kW d. 99.5 kW
Solution
To solve this problem, we first need to convert the speed from km/h to m/s.
1 km = 1000 m and 1 hour = 3600 seconds.
So, 42.0 km/h = 42.0 * 1000 / 3600 = 11.67 m/s.
Next, we calculate the work done by the truck using the formula Work = Force * Distance. In this case, the distance is the distance the truck travels in one second, which is its speed. So, Work = Force * Speed = 2.40 x 10^3 N * 11.67 m/s = 2.80 x 10^4 Joules.
This is the work done by the truck in one second, which is also the power in Watts. To convert to kilowatts, we divide by 1000, so the power is 2.80 x 10^4 / 1000 = 28.0 kW.
However, this is the power output, and we know that the truck is only running at 24.5% efficiency. So, the power input, or the power the motor must generate, is Power output / Efficiency = 28.0 kW / 0.245 = 114 kW.
So, the answer is c. 114 kW.
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