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?
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%.
To calculate the efficiency of the elevator, we can use the formula:
Efficiency = (Useful Energy Output / Total Energy Input) x 100%
In this case, the useful energy output is the work done by the elevator motor, which can be calculated using the formula:
Work = Force x Distance
The force can be found using Newton's second law of motion:
Force = mass x acceleration
First, let's calculate the force:
Given:
Mass = 2500 kg
We can assume the acceleration is constant throughout the lift. The acceleration can be found using the kinematic equation:
Distance = Initial Velocity x Time + (1/2) x acceleration x time^2
Since the initial velocity is 0 (stationary elevator), the equation simplifies to:
Distance = (1/2) x acceleration x time^2
Rearranging the equation, we get:
acceleration = (2 x Distance) / (time^2)
Given:
Distance = 12.0 m
Time = 3.00 s
Substituting the values, we find:
acceleration = (2 x 12.0 m) / (3.00 s)^2
= (24.0 m) / (9.00 s^2)
= 2.67 m/s^2
Now, let's calculate the force:
Force = mass x acceleration
= 2500 kg x 2.67 m/s^2
= 6675 N
Next, let's calculate the useful energy output (work):
Work = Force x Distance
= 6675 N x 12.0 m
= 80,100 J
Now, we can calculate the efficiency:
Efficiency = (Useful Energy Output / Total Energy Input) x 100%
= (80,100 J / 300,000 J) x 100%
= 26.7%
Therefore, the efficiency of the elevator is 26.7%.