Question

I would appreciate some help.

Answer 1

You oughtta be able to do this one.

The "efficiency" is just the portion of the input work that
comes out in a useful form.

If the efficiency is 70%, that tells you that however much work
you put INto the machine, the machine will do 70% of that much
work for you at the output side.

Put 20,000 J in ... out comes (0.70) x (20,000 J)  =  14,000 J .

What happens to the other 30% of the work you put into it ?
It turns into HEAT.  That's why machines always have to be
cooled somehow while they're running.