Mr. Franklin uses a machine that has an efficiency of 80%. The machine applies a force of 800 N through a distance of 20 m. Mr. Franklin applies a force of 50 N on the machine. Through what distance does Mr. Franklin apply the force? 4 m 16 m 260 m 400 m

The distance Mr Franklin applied the force is 400 m

Explanation:

The efficiency of the machine = 80%

The force applied by the machine = 800 N

The distance over which the force is applied by the machine = 20 m

The force applied by Mr. Franklin = 50 N

The work done by the machine = 800 × 20 = 16000 J

The ideal work at 100% efficiency = X

80% of X = 16000 J

X = 16000 N/0.8 = 20,000 J

Therefore, we have have;

Work = Force × Distance;

20,000 J = 50 N × Distance Mr Franklin applied the force

Distance Mr Franklin applied the force = 20,000 J/(50 N) = 400 m

The distance Mr Franklin applied the force = 400 m.

1/0.14nm = 10^10/1.4 a lot