Question

A machine of 75 percent efficiency is used to lift a load of 1 x 10^3 N through a height of 4.8 m?

Answer 1

What is the question?

Answer 2

(a) How much work is done on the load?

(b) How much energy is wasted?

Answer 3

w= what?

Answer 4

what is the equation for work?

Answer 5

i know already the question for a can you answer letter b?

Answer 6

w=fd is the equation

Answer 7

the change in kinetic energy is = work total, correct?

Answer 8

how about the wasted energy?

Answer 9

I am working on that. Lol

Answer 10

There are two types of energy, kinetic and potential.

Answer 11

work done lifting and lowering an object is w=-mgdcos(theta)

Answer 12

Work = (Force)(Distance) = (1000 N)(4.8 m) = 4,800 Joules

Energy wasted = (4,800 J)(25) / 75 = 1,600 Joules.

this is the answer i saw in other web. im confused! how to get the 25 there?

Answer 13

I am not sure, sorry.

Answer 14

Efficiency is \(\dfrac{W_\text{out}}{W_\text{in}}\), right? I want to use algebra, so I'll use \(\epsilon\) for efficiency. Then \(\dfrac{W_\text{out}}{W_\text{in}}=\epsilon\).

Input is how much the machine is given, and output is what the machine actually does.

So, we know efficiency, \(\epsilon=.75\).
And we know what the machine does, \(W_\text{out}=4,800\ \text J\).
So we can find how much work we put in.
\(W_\text{in}=\dfrac{W_\text{out}}{\epsilon}\)

And the energy wasted is the energy we put in but didn't get out. The input minus the output. So that is this, and I substitute the equation above for \(W_\text{in}\):
\(W_\text{in}-W_\text{out}=\dfrac{W_\text{out}}{\epsilon}-W_\text{out}\)

And this simplifies to \(W_\text{out}\left(\dfrac{1}{\epsilon}-1\right)\) in case you want that. And that gives the same answer, \(1,600\ \text J\).

Answer 15

I can use this formula to get to that person's, too.
\(W_\text{out}\left(\dfrac{1}{\epsilon}-1\right)\\
=W_\text{out}\left(\dfrac{1}{\epsilon}-\dfrac{\epsilon}{\epsilon}\right)\\
=W_\text{out}\left(\dfrac{1-\epsilon}{\epsilon}\right)
\)

Just more simplification.

Answer 16

thank you very much! your such a hero!

Answer 17

Haha, thanks! I'm glad I could help. Take care, and feel free to ask any questions.