Math Skills Kinetic Energy Answer Key - Fill Online, Printable ... - Free Printable

Let’s solve the practice problems step by step.

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Practice Problem 1:

> A 65.0 kg skydiver jumps from a plane. Her speed steadily increases until air resistance produces a force that balances the force due to her fall. How fast is the skydiver falling if her kinetic energy at this moment is 7.00 × 10⁵ J?

We are given:
- Mass (m) = 65.0 kg
- Kinetic Energy (KE) = 7.00 × 10⁵ J
- We need to find speed (v)

The formula for kinetic energy is:

KE = ½ mv²

We rearrange it to solve for v:

Multiply both sides by 2 →
2 × KE = m × v²

Divide both sides by m →
v² = (2 × KE) / m

Take square root of both sides →
v = √[(2 × KE) / m]

Now plug in the numbers:

v = √[(2 × 7.00 × 10⁵) / 65.0]
= √[(1,400,000) / 65.0]
= √[21,538.46...]
≈ 146.76 m/s

Rounded to 3 significant figures (since all inputs have 3):
→ 147 m/s

Wait — let’s double-check calculation:

2 × 7.00 × 10⁵ = 1,400,000
1,400,000 ÷ 65.0 = 21,538.4615...
√21,538.4615 ≈ ?

Let me calculate more precisely:

146² = 21,316
147² = 21,609
So between 146 and 147.

Try 146.7² = (146 + 0.7)² = 146² + 2×146×0.7 + 0.7² = 21,316 + 204.4 + 0.49 = 21,520.89
Still low.

146.8² = 146.7² + 2×146.7×0.1 + 0.01 ≈ 21,520.89 + 29.34 + 0.01 = 21,550.24 → too high.

Our target is 21,538.46

Try 146.75² = ?

(146.7 + 0.05)² = 146.7² + 2×146.7×0.05 + 0.0025
= 21,520.89 + 14.67 + 0.0025 = 21,535.5625

Close! Target is 21,538.46 → difference ~2.9

Add about 0.01 more: try 146.76²

= 146.75² + 2×146.75×0.01 + 0.0001 ≈ 21,535.5625 + 2.935 + 0.0001 ≈ 21,538.4976 → very close!

So v ≈ 146.76 m/s

But we must consider significant figures.

Given values:
- 65.0 kg → 3 sig figs
- 7.00 × 10⁵ J → 3 sig figs

So answer should be 3 sig figs → 147 m/s

✔ Confirmed.

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Practice Problem 2:

> The kinetic energy of a golf ball in midair is 143.3 J. If the golf ball has a mass of about 47 g, what is its speed?

First, convert mass to kilograms because SI unit for mass in physics formulas is kg.

47 g = 47 ÷ 1000 = 0.047 kg

Given:
- KE = 143.3 J
- m = 0.047 kg
- Find v

Again use:
KE = ½ mv²

Solve for v:

v = √[(2 × KE) / m]

Plug in:

v = √[(2 × 143.3) / 0.047]
= √[286.6 / 0.047]
= √[6097.8723...]

Calculate square root:

What’s √6097.87?

Note: 78² = 6084
79² = 6241

So between 78 and 79.

78.1² = 78² + 2×78×0.1 + 0.1² = 6084 + 15.6 + 0.01 = 6099.61 → a bit high

78.0² = 6084
Target: 6097.87 → difference = 13.87

Increment from 78.0: each 0.1 adds about 15.6 (from above), so 13.87 / 15.6 ≈ 0.89 → so approx 78.089

Try 78.08²:

= (78 + 0.08)² = 78² + 2×78×0.08 + 0.08² = 6084 + 12.48 + 0.0064 = 6096.4864

Still low.

78.09² = 78.08² + 2×78.08×0.01 + 0.0001 ≈ 6096.4864 + 1.5616 + 0.0001 = 6098.0481 → slightly over

Target: 6097.87 → so interpolate:

Between 78.08 and 78.09

Difference: 6098.0481 - 6096.4864 = 1.5617
We need 6097.87 - 6096.4864 = 1.3836

Fraction: 1.3836 / 1.5617 ≈ 0.886

So v ≈ 78.08 + 0.00886 ≈ 78.089 m/s

But let’s check with calculator-style precision:

286.6 ÷ 0.047 = 6097.87234...

√6097.87234 ≈ 78.0889...

So approximately 78.09 m/s

Now check significant figures:

- KE = 143.3 J → 4 sig figs
- Mass = 47 g → 2 sig figs (unless written as 47.0, but it says “about 47 g” → likely 2 sig figs)

In multiplication/division, the least number of sig figs determines the answer.

Mass has 2 sig figs → so answer should have 2 sig figs.

78.09 rounded to 2 sig figs → 78 m/s

Wait — 78 has two sig figs? Yes, if it’s exactly 78, it could be ambiguous, but since 78.09 rounds to 78 when using 2 sig figs? Actually, 78 has two sig figs only if we assume it’s not exact. But here, 78.09 to two sig figs is 78, because 78 is already two digits.

But let’s think: 47 g has two sig figs → uncertainty is ±1 g or so → relative error ~2%

So speed should reflect that.

78.09 m/s → with 2 sig figs → 78 m/s

But wait — 78 has two sig figs? In scientific notation, 7.8 × 10¹ would clearly show two sig figs.

However, 78 without decimal may be interpreted as having two sig figs if context implies it.

Alternatively, perhaps the problem expects us to treat 47 g as exact? But it says “about 47 g”, which suggests approximation → 2 sig figs.

But let’s see the example in the worksheet: they used 2.00 × 10³ kg (3 sig figs) and got 28.8 m/s (3 sig figs).

Here, 143.3 has 4, 47 has 2 → so answer should have 2.

But 78.09 to 2 sig figs is 78 — but 78 has two sig figs? Technically yes, if we write it as 78, it can be considered to have two.

To be safe, let’s express as 7.8 × 10¹ m/s → which clearly shows two sig figs.

But the question doesn’t specify format. Since the first problem gave answer as 28.8 (three sig figs), and here mass is “about 47 g”, I think 78 m/s is acceptable.

But let’s recalculate with exact values:

v = sqrt(2 * 143.3 / 0.047) = sqrt(286.6 / 0.047) = sqrt(6097.87234) = 78.0889...

If we take mass as 47.0 g (implying 3 sig figs), then 0.0470 kg → then answer would be 78.1 m/s (3 sig figs).

But the problem says “about 47 g” — which typically means 2 sig figs.

Looking back at the solved example: they used 2.00 × 10³ kg — three sig figs — and answer was 28.8 — three sig figs.

Here, 143.3 has four, 47 has two → so limiting is two.

Thus, 78 m/s

But 78.0889 rounded to two sig figs: look at the third digit — 8.0889 → first two digits are 7 and 8, next is 0 (in 78.0), so actually 78 has two sig figs, and since the next digit is 0 (<5), we keep 78.

Wait — no: 78.0889 — to two sig figs: start from left: 7 and 8 are the first two significant digits. The next digit is 0 (tenths place), which is less than 5, so we round down → still 78.

But 78 what? 78 m/s.

However, in many textbooks, if the number is between 10 and 100, and you want two sig figs, you might write it as 78, implying two.

To avoid ambiguity, sometimes they expect scientific notation.

But for this level, I think 78 m/s is fine.

Wait — let me check the calculation again with precise values:

2 * 143.3 = 286.6

286.6 / 0.047 = let's do division:

0.047 * 6097.87234 = ? Better to compute directly:

286.6 ÷ 0.047 = 286600 ÷ 47 (multiply numerator and denominator by 1000)

286600 ÷ 47:

47 * 6000 = 282,000

Subtract: 286,600 - 282,000 = 4,600

47 * 97 = 47*100 - 47*3 = 4700 - 141 = 4559

4600 - 4559 = 41

So 6097 + 41/47 ≈ 6097.87234 — same as before.

sqrt(6097.87234) = ?

Use better method: 78.09^2 = 78.09 * 78.09

78 * 78 = 6084

78 * 0.09 = 7.02, doubled is 14.04 (for cross terms)

0.09 * 0.09 = 0.0081

Better: (a+b)^2 = a^2 + 2ab + b^2, a=78, b=0.09

a^2 = 6084

2ab = 2*78*0.09 = 14.04

b^2 = 0.0081

Sum: 6084 + 14.04 = 6098.04 + 0.0081 = 6098.0481 — as before.

Our value is 6097.87234, which is less.

Difference: 6098.0481 - 6097.87234 = 0.17576

Derivative: d(v^2)/dv = 2v, so dv = d(v^2)/(2v) ≈ -0.17576 / (2*78.09) ≈ -0.17576 / 156.18 ≈ -0.001125

So v ≈ 78.09 - 0.001125 ≈ 78.088875 — matches earlier.

So v = 78.089 m/s

With 2 sig figs: 78 m/s

But let's see if the problem intends 47 g as 47.0 — probably not, since it says "about".

Perhaps in context of the worksheet, they expect 3 sig figs.

Look at the solved example: mass 2.00e3 kg (3 sig figs), KE 8.00e5 J (3 sig figs), answer 28.8 m/s (3 sig figs).

Here, KE is 143.3 (4 sig figs), mass 47 g — if we interpret as 47.0 g, then 3 sig figs.

"About 47 g" might mean approximately 47, so 2 sig figs, but in many school problems, they might expect you to use the numbers as given without worrying too much.

To match the style of the worksheet, which uses 3 sig figs in answers, and since 143.3 has 4, perhaps they want 3 sig figs for answer.

47 g — if it's 47, it could be 2, but let's calculate with 47.0 g = 0.0470 kg

Then v = sqrt(2*143.3 / 0.0470) = sqrt(286.6 / 0.0470) = sqrt(6097.87234) same as before? No:

0.0470 is same as 0.047 numerically, but sig figs different.

Numerically same calculation.

But if mass is 0.0470 kg (3 sig figs), then answer should have 3 sig figs.

78.089 → to 3 sig figs is 78.1 m/s

Because 78.089, the third digit is 0 (hundredths? Let's see: 78.089 — digits: 7,8,0,8,9 — first three significant digits are 7,8,0 — but 78.0 has three sig figs.

78.089 rounded to three sig figs: look at the fourth digit, which is 8 >5, so round up the third digit from 0 to 1? No.

Significant figures for 78.089:

- All non-zero digits are significant.
- Zeros between non-zero are significant.
- Leading zeros not significant.
- Trailing zeros after decimal are significant.

So 78.089 has five sig figs.

To round to three sig figs: look at the first three digits: 7,8,0 — and the next digit is 8, which is greater than 5, so we round up the last retained digit.

The number is 78.089

The first three significant digits are 7,8, and the next is 0 (which is the tenths place).

Actually, 78.089 — the digit in the units place is 8, tens is 7, tenths is 0, hundredths is 8, etc.

To three significant figures: we want to keep up to the tenths place if necessary.

Standard way: 78.089

- The first significant figure is 7 (tens)
- Second is 8 (units)
- Third is 0 (tenths)
- Fourth is 8 (hundredths) — which is >=5, so we round up the third digit.

Third digit is 0, rounding up makes it 1, but since it's 0, and we add 1, it becomes 1, and carry over? No.

78.089 — the third significant digit is the 0 in the tenths place.

Rounding up the 0 by 1 makes it 1, so 78.1

Yes: 78.089 rounded to three significant figures is 78.1

Because 78.089 is closer to 78.1 than to 78.0? 78.1 - 78.089 = 0.011, 78.089 - 78.0 = 0.089, so yes, closer to 78.1.

And since the digit after the third sig fig is 8>5, we round up.

So if we take mass as having 3 sig figs (47.0 g), then answer is 78.1 m/s.

Given that the worksheet example used 3 sig figs, and "about 47 g" might be intended as 47.0, I'll go with 78.1 m/s for consistency.

In many educational contexts, they don't strictly enforce sig figs for "about", but to match the style, 78.1 m/s is reasonable.

Let me confirm with the calculation:

v = sqrt(2 * 143.3 / 0.047) = sqrt(286.6 / 0.047)

If 0.047 is exact, then 286.6 / 0.047 = 6097.87234, sqrt=78.0889, which to 4 sig figs is 78.09, but since KE has 4, mass has 2, it's messy.

Perhaps the problem expects us to use the numbers as is and report based on input.

I recall that in the solved example, they had KE = 8.00e5 (3 sig figs), mass 2.00e3 (3 sig figs), answer 28.8 (3 sig figs).

Here, for problem 2, KE=143.3 (4 sig figs), mass=47 g — if we consider 47 as 2 sig figs, answer should be 2 sig figs.

But 78 m/s is 2 sig figs.

However, 78 m/s might be written as 78, but in the context, perhaps they want the calculation as is.

Another way: let's see what the worksheet does. In the solution, they wrote 28.8 m/s, which is three digits.

For consistency, and since 143.3 has four, perhaps they intend mass to be taken as 47.0.

I think it's safer to provide 78.1 m/s, assuming 3 sig figs.

Or perhaps calculate exactly and see.

Let's do the math without sig fig worry first.

v = sqrt(2 * 143.3 / 0.047) = sqrt(286.6 / 0.047)

286.6 ÷ 0.047 = 6097.872340425532

sqrt(6097.872340425532) = 78.088875 (approximately)

So 78.09 m/s if rounded to 4 sig figs, but for practical purposes, 78.1 m/s is fine.

I think for the answer, I'll put 78.1 m/s, as it matches the precision of the worksheet examples.

So final answers:

Problem 1: 147 m/s

Problem 2: 78.1 m/s

But let's box them.

Final Answer:
147 m/s
78.1 m/s