Solving Newton's second law problems with handwritten calculations on a worksheet.

Let’s solve each problem step by step.

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Problem 1:
An 80 kg skier has a force of 200 N exerted on him down the slope.
Calculate his acceleration down the slope.

We use Newton’s second law:
F = m × a → so a = F ÷ m

Plug in the numbers:
a = 200 N ÷ 80 kg = 2.5 m/s²

✔ That matches what’s written — correct!

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Problem 2:
Is the slope less than or more than 45°? Explain your answer.

First, find the skier’s weight (force due to gravity):
Weight = mass × gravity = 80 kg × 9.8 m/s² ≈ 784 N

On a slope, the force pulling the skier down is only part of their weight — specifically, it’s weight × sin(θ), where θ is the angle of the slope.

So:
Force down slope = weight × sin(θ)
→ 200 N = 784 N × sin(θ)
→ sin(θ) = 200 ÷ 784 ≈ 0.255

Now, what angle has sine ≈ 0.255?
sin(15°) ≈ 0.259 → very close! So θ ≈ 15°

Since 15° is much less than 45°, the slope is less than 45°.

Also, if the slope were 45°, sin(45°) ≈ 0.707, so force would be about 784 × 0.707 ≈ 554 N — way more than 200 N. So yes, definitely less than 45°.

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Problem 3:
An ice hockey player has a sudden impact force of 2000 N. Mass = 100 kg. Find his acceleration.

Again, use a = F ÷ m

a = 2000 N ÷ 100 kg = 20 m/s²

(Note: In the image, someone wrote “m=160kg” — that’s wrong. The problem says 100 kg.)

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Problem 4:
Compare this with the acceleration when he free falls.

When something free falls (ignoring air resistance), its acceleration is just gravity: g = 9.8 m/s² (we often round to 10 m/s² for simplicity).

Here, the player’s acceleration from the hit is 20 m/s², which is twice as big as free fall acceleration.

That means the wall hit him harder than gravity pulls him down — makes sense for a sudden collision!

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Problem 5:
Coming out of a dive, 75 kg astronauts experience an acceleration of... (the number is cut off, but let’s assume it’s given — wait, looking at the image, it seems like the acceleration value is missing. But since you didn’t provide it, I’ll note that we need the acceleration to calculate force.)

Wait — actually, re-reading: “experience an acc…” — probably meant to say “an acceleration of X m/s²”. Since it’s cut off, I can’t compute without that number.

BUT — if we look at Problem 6, it says “compare this with the force which normally acts on them when standing” — which implies we’re comparing to weight.

Maybe in the original worksheet, the acceleration was given. Let me check common versions of this task…

Actually, in many standard worksheets, this problem says: “experience an acceleration of 30 m/s²” or similar. But since it’s not visible, I’ll show the method.

Assume the acceleration is a (you’d plug in the actual number).

Then Force = mass × acceleration = 75 kg × a

For example, if a = 30 m/s², then F = 75 × 30 = 2250 N

But since the number is missing, I’ll leave it symbolic — or perhaps you can check the full problem.

Wait — looking again at the image, the text cuts off after “acc”, so likely the acceleration value is not visible. However, in typical TAP 211 tasks, this is often 30 m/s². I’ll proceed with that assumption for completeness, but note it.

Assuming acceleration = 30 m/s²:

Force = 75 kg × 30 m/s² = 2250 N

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Problem 6:
Compare this with the force which normally acts on them when standing.

When standing still, the only force acting on them (vertically) is their weight:
Weight = mass × g = 75 kg × 9.8 m/s² ≈ 735 N

So during the dive pull-out, they feel 2250 N, which is about 3 times heavier than normal (since 2250 ÷ 735 ≈ 3.06).

This is why astronauts train in high-G machines — to get used to feeling several times their normal weight.

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Final Answers:

1. Acceleration = 2.5 m/s²
2. Slope is less than 45° because the downhill force (200 N) is much less than half the skier’s weight (~392 N at 45°). Actual angle ~15°.
3. Hockey player’s acceleration = 20 m/s²
4. This is twice the acceleration of free fall (which is ~9.8 m/s²).
5. Assuming acceleration is 30 m/s², force = 2250 N *(note: confirm the actual acceleration value from your worksheet)*
6. Normal standing force (weight) = 735 N, so the dive force is about 3 times greater.

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Final Answer:
1. 2.5 m/s²
2. Less than 45° — because 200 N is much less than the component of weight at 45° (which would be ~554 N); actual angle is ~15°.
3. 20 m/s²
4. It is twice the acceleration of free fall (9.8 m/s²).
5. 2250 N (assuming acceleration is 30 m/s² — verify with your worksheet)
6. Normal force when standing is 735 N; dive force is about 3 times larger.