an elevator motor lifts a load of 2500 kg a height of 12.0 m in 3.00 seconds. the energy required by the motor to accomplish this task is 300,000 J. what is the efficiency of the elevator?

3 answers

The efficiency of a system is defined as the ratio of useful output energy to the input energy.

In this case, the useful output energy is the work done by the elevator motor in lifting the load, which is given as 300,000 J.

The input energy is the total energy consumed by the elevator motor.

The total energy consumed by the motor can be calculated as the product of the power (energy per unit time) and the time taken.

Since the power is the work done (energy) divided by the time, we can write:

Power = Work/Time

Plugging in the given values, we have:

Power = 300,000 J / 3.00 s = 100,000 W

Now, we can calculate the total energy consumed:

Energy consumed = Power * Time

Energy consumed = 100,000 W * 3.00 s = 300,000 J

Comparing this value with the given input energy, we find that the total energy consumed by the motor is also 300,000 J.

Therefore, the efficiency of the elevator can be calculated as the ratio of useful output energy to the input energy:

Efficiency = Useful output energy / Input energy

Efficiency = 300,000 J / 300,000 J = 1

Therefore, the efficiency of the elevator is 1 or 100%.
NO again !

"..... lifts a load of 2500 kg a height of 12.0 m ......"
Lift force = m g = 2500 * 9.8 = about 24,400 Newtons
so work done = 24, 400 Newtons * 12 meters
work in = 300, 000
eff = 100 * (24,400 *12 / 300,000)= 100 * (24.4 /25)
Apologies for the earlier mistake. Let's recalculate the efficiency correctly.

Given:
Mass of the load (m) = 2500 kg
Height (h) = 12.0 m
Work done by the motor (W) = 300,000 J

We know that work done is equal to force multiplied by distance. In this case, the force is the weight of the load, which is the mass multiplied by the acceleration due to gravity.

Weight (W) = mass (m) * acceleration due to gravity (g)
Weight (W) = 2500 kg * 9.8 m/s^2
Weight (W) = 24,500 N

Now, we can calculate the work done:
Work (W) = force (F) * distance (d)
Work (W) = 24,500 N * 12.0 m
Work (W) = 294,000 J

Now, we can calculate the efficiency:
Efficiency = useful output energy / input energy
Efficiency = (work done / input energy) * 100
Efficiency = (294,000 J / 300,000 J) * 100
Efficiency = 0.98 * 100
Efficiency = 98%

Therefore, the efficiency of the elevator is 98%.