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I have a simpler explanation. Energy dissipated = energy wasted. So, using the formula of efficiency, the total energy used by the electric motor is 72kJ. Since the truck only gained 48kJ, wasted energy = 72-48 = 24kJ(B).Aaryan34532 said:Homework Statement
See picture.View attachment 234970
Homework Equations
The Attempt at a Solution
I tried at first just doing (1/3)*(48kJ) to get the energy dissipated, but that would i=give me 16kJ
The efficiency of an electric motor directly impacts the overall performance of pulling a truck up an inclined plane. A higher efficiency motor will require less energy and therefore perform better in terms of speed and power.
The efficiency of an electric motor pulling a truck up an inclined plane can be influenced by several factors, including the type and quality of the motor, the weight and size of the truck, and the angle of the incline.
While a higher horsepower motor may seem like the more efficient choice, it is not always the case. In some scenarios, a lower horsepower motor with a higher efficiency may be more suitable for pulling a truck up an inclined plane.
The efficiency of an electric motor can be improved for pulling a truck up an inclined plane by using a motor with a higher power-to-weight ratio, reducing friction between the motor and the truck, and selecting the appropriate motor for the specific task.
Yes, there are alternative methods for pulling a truck up an inclined plane that may be more efficient than using an electric motor. These include using a pulley system or employing a combination of human and mechanical power.