Question

Because there is very little friction, the lever is an extremely efficient simple machine. using a 90.0% efficient lever, what input work is required to lift a(n) 19.0 kg mass through a distance of 45 cm? 3232 incorrect: your answer is incorrect. seenkey 93.2 j

Answer 1

more simply efficiency n=output work/input work =mgh/W(in)W(in) =18*9.81*0.5/0.90=98.1 Joule

Answer 2

Input work: 93.2 J

Step-by-step explanation:

The work efficiency formula is:

efficiency = output work/ input work

Solving for input work:

input work = output work/efficiency

The work required to lift an object is computed as follows:

Output work = mass * gravity * height

Using mass in kg, gravity in m/(s^2) and height in m; then work is expressed in Joules. Combining the equations and replacing with data, we get (units are omitted):

input work = 19*9.81*0.45/0.9 = 93.2 J

Note that 45 cm is equivalent to 0.45 m