An electric motor is tested and found 76% efficient. if the input work is 400 J, what is the motors output work?

A ramp has a length of 4m and a height of 1.5m. If a man pushes a 40N crate to the top of the ramp with a force of 60N, what is the efficiency of the ramp?

How do I do these two?

To find the output work of an electric motor, we need to use the efficiency formula:

Efficiency = (Output work / Input work) * 100%

Given that the efficiency of the motor is 76% and the input work is 400 J, we can rearrange the formula to solve for output work:

Output work = (Efficiency / 100%) * Input work

Substituting the given values:

Output work = (76 / 100) * 400

Output work = 0.76 * 400

Output work = 304 J

Therefore, the output work of the electric motor is 304 J.

Now let's move on to the efficiency of the ramp. In this case, efficiency is calculated as the ratio of useful output work to input work:

Efficiency = (Output work / Input work) * 100%

The given information states that the man pushes the crate with a force of 60N, so the useful output work can be calculated using:

Useful output work = Force * Distance

The distance moved along the ramp can be calculated using the pythagorean theorem:

Distance = √(length^2 + height^2)

Substituting the given values:

Distance = √(4^2 + 1.5^2)

Distance = √(16 + 2.25)

Distance = √18.25

Distance ≈ 4.27 m

Now we can calculate the useful output work:

Useful output work = 60N * 4.27m

Useful output work ≈ 256.2 J

Substituting the values into the efficiency formula:

Efficiency = (256.2 J / Input work) * 100%

It seems that the question did not provide the input work for the ramp. Without that information, the efficiency calculation cannot be determined.