An electric hoist is used to lift a 250.0 kg load to a height of 80.0 m in 98.0 s. What is the power of the hoist motor? Group of answer choices 510 W 5,000 W 20,000 W 196,000 W 2,000 W
The correct answer and explanation is:
To find the power of the hoist motor, we use the definition of power in physics: Power=WorkTime\text{Power} = \frac{\text{Work}}{\text{Time}}
Step 1: Calculate the Work Done
The work done in lifting the object is equal to the gravitational potential energy gained: Work=Force×Distance\text{Work} = \text{Force} \times \text{Distance}
Here, the force required to lift the object is equal to its weight, which is: Force=mass×g=250.0 kg×9.8 m/s2=2450 N\text{Force} = \text{mass} \times g = 250.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 2450 \, \text{N}
The distance is the height the object is lifted: Distance=80.0 m\text{Distance} = 80.0 \, \text{m}
So, the work done is: Work=2450 N×80.0 m=196,000 Joules\text{Work} = 2450 \, \text{N} \times 80.0 \, \text{m} = 196,000 \, \text{Joules}
Step 2: Calculate the Power
Now divide the work by the time: Power=196,000 J98.0 s=2,000 W\text{Power} = \frac{196,000 \, \text{J}}{98.0 \, \text{s}} = 2,000 \, \text{W}
✅ Correct Answer: 2,000 W
Explanation (300 words):
Power is the rate at which work is done. In the context of machines like an electric hoist, it tells us how quickly the machine can transfer energy—in this case, to lift a load vertically against gravity. The hoist lifts a 250.0 kg load to a height of 80.0 m, meaning it does work equal to the increase in gravitational potential energy.
Gravitational potential energy is given by PE=mghPE = mgh, where mm is mass, gg is acceleration due to gravity, and hh is height. The mass of 250.0 kg and height of 80.0 m result in a potential energy gain of 196,000 joules. This energy is transferred in 98.0 seconds.
Power is calculated by dividing the work done (196,000 J) by the time taken (98.0 s), yielding 2,000 watts. This means the hoist motor outputs energy at a rate of 2,000 joules per second to lift the load. Understanding this helps engineers design appropriate motors and power supplies for lifting operations.
The power output also determines energy efficiency and safety. If the motor supplied less than this power, it wouldn’t be able to perform the lift in 98 seconds, or might stall. If it were more powerful, the lift could be quicker or the motor could handle heavier loads.
Therefore, the correct power required for the motor is 2,000 W.