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Difficulty: MediumWork, Energy and Power

An electric motor with an efficiency of 80%80\% is used to pull a 100 kg100\text{ kg} object up a smooth incline inclined at 30∘30^\circ to the horizontal at a constant speed of 2 m s−12\text{ m s}^{-1}. Taking acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}, what is the electrical power input to the motor, in watts?

Answer: 1250 W

Answer

1250 W
The force needed to move the mass up the smooth incline at constant speed is the parallel component of weight, F=mgsin⁡(30∘)=500 NF = mg \sin(30^\circ) = 500\text{ N}. The useful power output is Pout=Fv=500×2=1000 WP_{\text{out}} = Fv = 500 \times 2 = 1000\text{ W}. Dividing by the efficiency of 0.800.80 yields the electrical power input Pin=1250 WP_{\text{in}} = 1250\text{ W}.

Step-by-Step Solution

1
Determine the force required along the inclined plane.
F=mgsin⁡(30∘)=100 kg×10 m s−2×0.5=500 NF = mg \sin(30^\circ) = 100\text{ kg} \times 10\text{ m s}^{-2} \times 0.5 = 500\text{ N}
At constant velocity, the applied force balances the component of weight parallel to the incline.
2
Calculate the useful power output delivered by the motor.
Pout=F×v=500 N×2 m s−1=1000 WP_{\text{out}} = F \times v = 500\text{ N} \times 2\text{ m s}^{-1} = 1000\text{ W}
Mechanical power output is the product of pulling force and constant speed.
3
Calculate the total electrical power input required.
Pin=PoutEfficiency=1000 W0.80=1250 WP_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{1000\text{ W}}{0.80} = 1250\text{ W}
Efficiency is defined as the ratio of useful power output to total power input.

Key Concept

Mechanical power on inclined planes and system efficiency
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Topics

Work, Energy and Power

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