8th Grade Science worksheet focusing on mechanical advantage calculations for simple machines, featuring problems and diagrams.

Let’s solve each problem step by step using the formula:

Mechanical Advantage (MA) = Resistance Force (R) / Effort Force (E)
Or, if you’re given MA and need to find force:
Resistance Force = MA × Effort Force
Effort Force = Resistance Force / MA

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Problem 1:

Effort force = 30N
Resistance force (force on lid) = 90N
MA = R/E = 90 / 30 = 3

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Problem 2:

Rock mass = 800N → this is the resistance force (weight)
Effort force = 200N
MA = 800 / 200 = 4

*(Note: In science problems like this, when they say “mass of 800N”, they actually mean weight — which is a force. So we treat it as 800N force.)*

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Problem 3:

Effort force = 55N
Resistance force = 675N
MA = 675 / 55 ≈ 12.27 (we can leave it as a decimal or fraction)

But let’s calculate exactly:
675 ÷ 55 = 12.2727... → round to two decimals? Or keep as fraction?

Actually, 675 ÷ 55 = 135/11 ≈ 12.27

But since the worksheet doesn’t specify rounding, we’ll write it as 12.27 or maybe check if it simplifies.

Wait — 675 ÷ 5 = 135, 55 ÷ 5 = 11 → so 135/11 = 12.27...

We’ll go with 12.27 for now.

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Problem 4:

Effort force = 250N
Resistance force (box weight) = 1250N
MA = 1250 / 250 = 5

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Problem 5:

Effort force = 75N
MA = 5
Find resistance force (force applied to walnut)
Force = MA × Effort = 5 × 75 = 375N

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Problem 6:

MA = 6
Resistance force = 1800N
Find effort force
Effort = Resistance / MA = 1800 / 6 = 300N

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Problem 7:

Lever diagram:
Distance to Effort = 9m
Distance to Resistance = 3m
MA = Distance to Effort / Distance to Resistance = 9 / 3 = 3

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Problem 8:

Diagram shows total length = 20m, distance from fulcrum to load = 5m
So distance from fulcrum to effort = 20 - 5 = 15m
MA = Distance to Effort / Distance to Resistance = 15 / 5 = 3

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Now let’s double-check all answers:

1. 90/30 = 3 ✔
2. 800/200 = 4 ✔
3. 675/55 = 12.27 (or 135/11) — but let’s see if the worksheet expects exact value. Since others are whole numbers, maybe I made a mistake? Wait — 55 × 12 = 660, 675 - 660 = 15 → so 12 + 15/55 = 12 + 3/11 ≈ 12.27. It’s correct. We’ll write 12.27 or perhaps 12.3? But better to be precise. Actually, in school worksheets, sometimes they expect fractions. Let me check: 675 ÷ 55 = 135/11 — that’s exact. But since other answers are integers, maybe it’s okay to leave as decimal. I’ll use 12.27.

Wait — let me recalculate: 55 × 12.27 = 55×12 = 660, 55×0.27=14.85 → total 674.85 — close enough. Actually, 675 ÷ 55 = 12.2727..., so 12.27 is fine.

But looking at the context — maybe they want it rounded? Or perhaps I misread. No, the numbers are correct.

Alternatively, maybe the problem meant 675N output and 55N input — yes, so MA = 675/55 = 12.27. Okay.

4. 1250/250 = 5 ✔
5. 5 × 75 = 375 ✔
6. 1800/6 = 300 ✔
7. 9/3 = 3 ✔
8. 15/5 = 3 ✔

All good.

Final Answers:

1. MA = 3
2. MA = 4
3. MA = 12.27 (or 135/11 — but since worksheet uses decimals elsewhere, 12.27 is acceptable)
Wait — actually, let me check problem 3 again: "a force of 55N is applied... applies a force of 675N" — yes, so MA = 675/55.

But 675 ÷ 55 = 12.2727... — perhaps write as 12.3 if rounding to one decimal? But the worksheet doesn't specify. To be safe, I'll write the exact value as a decimal with two places: 12.27

However, in many school settings, they might expect you to simplify or notice if it's a nice number. 675 and 55 both divisible by 5: 135/11 — which is about 12.27. I think it's fine.

Alternatively, maybe I should write it as a mixed number? But the worksheet has blanks for MA = ___, so probably decimal is okay.

I’ll proceed.

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Final Answer:

1. MA = 3
2. MA = 4
3. MA = 12.27
4. MA = 5
5. Force = 375N
6. Force = 300N
7. MA = 3
8. MA = 3