Welcome Back

Sign in to your PART66Online account

One click — no password needed

or use email
Forgot password?

Don't have an account? Register here

How would you work out the work done by a machine assuming it is 100% efficient?

  • Velocity ratio and efficiency.
  • Mechanical advantage and output.
  • Mechanical advantage and input.

Explanation from PART66Online

At 100 per cent efficiency no energy is lost, so the work output equals the work input. Knowing the mechanical advantage (load over effort) together with the input effort and its movement lets you determine the load and hence the work done by the machine.

A

alexeft13 asked · 3 Oct 2020

WHY?

Community Comments (2)

D
Diverdeep 11 Apr 2023
instead of and there's should be multiplied by
A
Angel 25 Nov 2024
The confusion arises from the difference between calculating work done by the machine versus calculating the efficiency of the machine.

Your physics book's content focuses heavily on efficiency, and one of the ways to calculate it is indeed:

Efficiency=(Mechanical Advantage (MA)/Speed Ratio)×100

This formula links mechanical advantage to the input/output relationship and how well the machine minimizes losses. The book also emphasizes that no machine is 100% efficient, as losses always occur due to heat, friction, etc.

The Test Question
Your test question was specifically about work done by a machine, assuming it is 100% efficient. For this, the mechanical advantage and input are often used because:

Mechanical Advantage relates the force output to the force input:

Mechanical Advantage (MA)=Output Force/Input Force
or
Mechanical Advantage (MA)= Input Force/Output Force


To calculate work input, you need:

Work Input=Input Force×Distance Input.

When the machine is 100% efficient:

Work Output=Work Input.

Work done by the machine can then be derived using mechanical advantage (how the forces are transformed) and the input work provided.

Thus, "Mechanical Advantage and Input" is correct if you're focused on determining the relationship between the forces and work in a theoretical, ideal (100% efficient) scenario.

Why the Two Answers Differ
The physics book emphasizes real-world efficiency (with inevitable losses), while the test assumes an idealized scenario (no losses). This difference changes the approach:

For calculating efficiency: You use both output and input parameters (e.g., measured performance and ideal performance).
For determining work in a 100% efficient machine: You relate input (force/distance) and mechanical advantage directly, which is why the test prioritizes "Mechanical Advantage and Input."
The key lies in understanding the context of the question:

The test assumes an idealized situation (100% efficiency).
The book explains real-world machines, where efficiency is less than 100%.





Please Sign In to post a comment.

We use essential cookies to keep you signed in, plus anonymous analytics to understand how the site is used. Cookie-based analytics is set only with your consent. See our Privacy & Cookie Policy.