A motor that uses electrical energy at a rate of 1.5 kW is connected to a hydraulic lift that is used to lift a 1300 kg car to a height of 1.8 m in 24 s at a constant speed. Find the efficiency of the system consisting of the motor and the lift.

Answers

Answer 1
Answer:

Final answer:

To find the efficiency of the system, calculate the input power and output power and then divide the output power by the input power and multiply by 100%.

Explanation:

To find the efficiency of the system consisting of the motor and the lift, we need to calculate the input power and the output power. The input power is given as 1.5 kW. The output power can be calculated by multiplying the weight of the car by the height it was lifted and dividing by the time taken. The efficiency is then calculated by dividing the output power by the input power and multiplying by 100%.

Step 1: Calculate the output power:
Output Power = (Weight of the car) x (Height lifted) / (Time taken)

Step 2: Calculate the efficiency:
Efficiency = (Output Power / Input Power) x 100%

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Answer 2
Answer:

Final answer:

The efficiency of the hydraulic lift system is found to be 63.7% by comparing the useful power output of lifting the car with the electrical energy input of the motor.

Explanation:

To calculate the efficiency of the system consisting of the motor and the hydraulic lift, we need to compare the useful work output to the energy input. The work done in lifting the car is equal to the product of the gravitational force (weight of the car) and the height raised. With a 1300 kg car, this is 1300 kg × 9.8 m/s2 × 1.8 m. This comes out to 22,932 joules (since work is measured in joules). The time taken is 24 seconds.

Since power is the rate at which work is done or energy is transferred, the useful power output is the work done divided by the time, which is 22,932 joules / 24 seconds = 955.5 watts or 0.9555 kW. Meanwhile, the motor uses electrical energy at a rate of 1.5 kW. To find the efficiency, we divide the useful power by the power input and multiply by 100 to get a percentage:

Efficiency = (0.9555 kW / 1.5 kW) × 100 = 63.7%

Therefore, the efficiency of the motor and the hydraulic lift system is 63.7%.

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Answers

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