Question

A machine has a velocity ratio of 5 and a mechanical advantage of 4. The system is used to raise a load of 500N through 10m. The wasted energy is:

A) 6250 J
B) 5000 J
C) 1250 J
D) 12500 J

Answer 1

The wasted energy in the system is -4750 Joules.

Step-by-step explanation:

The wasted energy can be calculated by subtracting the useful work done from the total work done. The useful work done is equal to the load lifted multiplied by the distance lifted, which is 500N * 10m = 5000 J. The total work done can be calculated by multiplying the input force (effort force) by the input distance (effort distance). The input force can be calculated by dividing the load force by the mechanical advantage, which is 500N / 4 = 125 N. The input distance is equal to the output distance divided by the velocity ratio, which is 10m / 5 = 2m. Hence, the input distance is 2m. The total work done is then 125N * 2m = 250J. Therefore, the wasted energy is the difference between the total work done and the useful work done, which is 250J - 5000J = -4750J.