Physics problem set covering topics such as energy conservation, spring forces, and Newton's laws.

Let's solve Problem 14 step by step, as it is the first and most detailed one in your list.

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Problem 14:

A shotputter releases a 5.00 kg shot, which is 3.00 m above the ground and traveling at 15.0 m/s. It reaches a maximum height of 14.5 m above the ground and then falls to the ground. Air resistance is negligible.

We are asked to find:

(a) Potential energy of the shot as it left the hand relative to the ground

Use the formula for gravitational potential energy:
\[
U = mgh
\]
Where:
- \( m = 5.00 \, \text{kg} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( h = 3.00 \, \text{m} \)

\[
U = (5.00)(9.8)(3.00) = 147 \, \text{J}
\]

✔ Answer (a): 147 J

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(b) Kinetic energy of the shot as it left the hand

Use kinetic energy formula:
\[
K = \frac{1}{2}mv^2
\]
- \( v = 15.0 \, \text{m/s} \)

\[
K = \frac{1}{2}(5.00)(15.0)^2 = \frac{1}{2}(5.00)(225) = 562.5 \, \text{J}
\]

✔ Answer (b): 562.5 J

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(c) Total energy of the shot as it left the hand

Total mechanical energy (since no air resistance):
\[
E_{\text{total}} = K + U = 562.5 + 147 = 709.5 \, \text{J}
\]

✔ Answer (c): 709.5 J

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(d) Total energy of the shot as it reached its maximum height

Since air resistance is negligible, total mechanical energy is conserved.

So, total energy remains the same:
\[
E_{\text{total}} = 709.5 \, \text{J}
\]

✔ Answer (d): 709.5 J

---

(e) Potential energy of the shot at its maximum height

At max height: \( h = 14.5 \, \text{m} \)

\[
U = mgh = (5.00)(9.8)(14.5) = 710.5 \, \text{J}
\]

Wait — this seems higher than the total energy? That can't be!

Let’s double-check.

Wait! The total energy was 709.5 J, but we’re getting 710.5 J here? That suggests an inconsistency.

But let's compute carefully:

\[
(5.00)(9.8)(14.5) = 5 \times 9.8 \times 14.5
\]
\[
= 49 \times 14.5 = 710.5 \, \text{J}
\]

But earlier total energy was only 709.5 J — that’s a problem.

Ah! So something must be wrong.

Let’s go back.

Wait — maybe we made a mistake in (b)?

Wait: initial speed is 15.0 m/s, mass 5.00 kg

\[
K = \frac{1}{2}mv^2 = \frac{1}{2}(5)(225) = 562.5 \, \text{J} \quad \text{✓}
\]

Initial PE: \( mgh = 5 \times 9.8 \times 3 = 147 \, \text{J} \quad \text{✓} \)

Total: \( 562.5 + 147 = 709.5 \, \text{J} \)

But at max height, all kinetic energy should have converted to potential energy, so:

\[
U_{\text{max}} = E_{\text{total}} = 709.5 \, \text{J}
\]

But if we calculate using \( mgh \), with \( h = 14.5 \, \text{m} \):

\[
U = 5 \times 9.8 \times 14.5 = 710.5 \, \text{J}
\]

This is higher than total energy — impossible.

So contradiction.

That means either:
- The given max height of 14.5 m is inconsistent with the initial velocity?
- Or we need to check if the motion is vertical or angled?

But the problem says it "reaches a maximum height of 14.5 m", and initial speed is 15.0 m/s upward? But wait — is it thrown vertically?

It doesn’t say. But if it’s thrown at an angle, the vertical component matters.

But here’s the key: if the shot is released at 3.00 m height and reaches 14.5 m, then vertical displacement is \( \Delta h = 14.5 - 3.00 = 11.5 \, \text{m} \)

So, from conservation of energy:

At release:
\( E_i = \frac{1}{2}mv^2 + mg h_i \)

At max height:
\( E_f = mg h_f \) (since \( v_y = 0 \), kinetic energy due to vertical motion is zero; but if there's horizontal component, some KE remains!)

Ah! Here’s the issue: if the shot is thrown at an angle, it still has horizontal velocity at the top, so kinetic energy is not zero.

Therefore, potential energy at max height ≠ total energy, because kinetic energy is still present (horizontal).

So our earlier assumption is flawed.

But the problem says “reaches a maximum height of 14.5 m” — that implies the vertical component of velocity becomes zero at that point.

But the total kinetic energy at that point is not zero, unless it’s thrown straight up.

So unless specified otherwise, we assume it may have been thrown at an angle.

But the problem gives us both:
- Initial speed: 15.0 m/s
- Initial height: 3.00 m
- Max height: 14.5 m

So we can use this to determine what happens.

Let’s re-analyze.

Let’s suppose the shot is thrown with speed \( v_0 = 15.0 \, \text{m/s} \) at some angle, and it rises to a height of 14.5 m.

The vertical component of velocity can be found using:

\[
v_y^2 = v_{y0}^2 - 2g\Delta h
\]

At max height, \( v_y = 0 \), so:

\[
0 = v_{y0}^2 - 2g(14.5 - 3.00) = v_{y0}^2 - 2(9.8)(11.5)
\]
\[
v_{y0}^2 = 2 \times 9.8 \times 11.5 = 225.4
\]
\[
v_{y0} = \sqrt{225.4} \approx 15.01 \, \text{m/s}
\]

But initial speed is only 15.0 m/s, so vertical component cannot exceed 15.0 m/s.

But here \( v_{y0} \approx 15.01 \), which is slightly more than 15.0 — impossible.

So contradiction.

Wait: 2 × 9.8 × 11.5 = ?

Compute:
\[
2 \times 9.8 = 19.6
\]
\[
19.6 × 11.5 = 19.6 × 10 + 19.6 × 1.5 = 196 + 29.4 = 225.4
\]
Yes.

So \( v_{y0}^2 = 225.4 \Rightarrow v_{y0} = \sqrt{225.4} \approx 15.01 \, \text{m/s} \)

But total speed is 15.0 m/s, so vertical component cannot be greater than 15.0.

So impossible to reach 14.5 m from 3.00 m with only 15.0 m/s speed.

Wait — unless gravity is taken as 10 m/s²? Let's try that.

Try \( g = 10 \, \text{m/s}^2 \):

Then:
\[
v_{y0}^2 = 2×10×11.5 = 230 \Rightarrow v_{y0} = \sqrt{230} \approx 15.17 \, \text{m/s}
\] — even worse.

Still impossible.

Wait — maybe the initial speed is not purely vertical, but the maximum height is 14.5 m above ground, so Δh = 11.5 m.

But with initial speed 15.0 m/s, what’s the maximum possible height?

If thrown straight up:
\[
v^2 = v_0^2 - 2gΔh \Rightarrow 0 = (15.0)^2 - 2×9.8×Δh
\]
\[
Δh = \frac{225}{19.6} \approx 11.48 \, \text{m}
\]

So maximum height above release point is about 11.48 m → total height = 3.00 + 11.48 = 14.48 m ≈ 14.5 m

Ah! So it’s consistent.

So yes, the shot was likely thrown straight up (or very close), and the 14.5 m is approximately correct.

So we can proceed.

So:

Maximum height above ground: 14.5 m

So potential energy at max height:

\[
U = mgh = 5.00 × 9.8 × 14.5 = ?
\]

Calculate:
\[
5 × 9.8 = 49
\]
\[
49 × 14.5 = 49 × (14 + 0.5) = 49×14 = 686, 49×0.5 = 24.5 → total = 710.5 \, \text{J}
\]

But total energy was only 709.5 J — off by 1 J.

Why?

Because when thrown straight up, the max height is:

\[
h_{\text{max}} = h_0 + \frac{v_0^2}{2g} = 3.00 + \frac{(15.0)^2}{2×9.8} = 3.00 + \frac{225}{19.6} = 3.00 + 11.4796 ≈ 14.4796 \, \text{m}
\]

So actual max height is 14.48 m, not 14.5 m.

But problem says 14.5 m — probably rounded.

So we can accept it as approximate.

Alternatively, perhaps they expect us to use the given values and ignore the slight inconsistency.

But better to use conservation of energy.

So let’s do it properly.

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Let’s re-solve with consistency.

Given:

- Mass \( m = 5.00 \, \text{kg} \)
- Initial height \( h_i = 3.00 \, \text{m} \)
- Initial speed \( v_i = 15.0 \, \text{m/s} \)
- Max height \( h_{\text{max}} = 14.5 \, \text{m} \)

Assume no air resistance → mechanical energy conserved.

(a) Potential energy at release:

\[
U_i = mgh_i = 5.00 × 9.8 × 3.00 = 147 \, \text{J}
\]

✔ Answer (a): 147 J

(b) Kinetic energy at release:

\[
K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(5.00)(15.0)^2 = 0.5 × 5 × 225 = 562.5 \, \text{J}
\]

✔ Answer (b): 562.5 J

(c) Total energy at release:

\[
E_i = K_i + U_i = 562.5 + 147 = 709.5 \, \text{J}
\]

✔ Answer (c): 709.5 J

(d) Total energy at max height:

Conserved → 709.5 J

✔ Answer (d): 709.5 J

(e) Potential energy at max height:

\[
U_{\text{max}} = mgh_{\text{max}} = 5.00 × 9.8 × 14.5 = 710.5 \, \text{J}
\]

But this is greater than total energy (709.5 J) — impossible.

So contradiction.

But wait — unless the speed is not 15.0 m/s, or height is not 14.5 m, or g is different.

Wait — perhaps the problem expects us to use g = 10 m/s²?

Try with \( g = 10 \):

(a) \( U_i = 5 × 10 × 3 = 150 \, \text{J} \)

(b) \( K_i = 0.5 × 5 × 225 = 562.5 \, \text{J} \)

(c) Total = 150 + 562.5 = 712.5 J

(e) \( U_{\text{max}} = 5 × 10 × 14.5 = 725 \, \text{J} \) — even worse.

No.

Wait — maybe the initial speed is not 15.0 m/s, but rather the vertical component?

But the problem says “traveling at 15.0 m/s” — so total speed.

But if it’s thrown at an angle, the vertical component is less than 15.0.

But then how can it reach 14.5 m?

Let’s compute the required vertical speed to reach 14.5 m from 3.00 m:

\[
\Delta h = 11.5 \, \text{m}
\]

\[
v_{y0}^2 = 2g\Delta h = 2 × 9.8 × 11.5 = 225.4
\Rightarrow v_{y0} = \sqrt{225.4} ≈ 15.01 \, \text{m/s}
\]

But total speed is only 15.0 m/s, so vertical component cannot exceed 15.0 m/s.

So impossible to reach 14.5 m.

But if we use \( g = 9.8 \), and \( v_0 = 15.0 \), then max height above release is:

\[
\Delta h = \frac{v_0^2}{2g} = \frac{225}{19.6} ≈ 11.48 \, \text{m}
\]

So max height above ground = 3.00 + 11.48 = 14.48 m, which rounds to 14.5 m.

So the value 14.5 m is rounded.

So we should use 14.48 m or accept 14.5 m as approximation.

But for energy calculations, better to use conservation of energy.

So at max height, kinetic energy is not zero if it’s not thrown straight up — but if it is thrown straight up, then at max height, velocity is zero, so kinetic energy is zero.

But if thrown straight up, then:

- Initial KE = 562.5 J
- Initial PE = 147 J
- Total = 709.5 J

At max height:
- PE = mgH = 5 × 9.8 × (3 + 11.48) = 5 × 9.8 × 14.48

Calculate:
\[
5 × 9.8 = 49
\]
\[
49 × 14.48 = ?
\]
\[
49 × 14 = 686
\]
\[
49 × 0.48 = 23.52
\]
Total = 686 + 23.52 = 709.52 J ≈ 709.5 J

Perfect.

So at max height, PE = 709.5 J, KE = 0

But the problem says max height is 14.5 m — so let’s use that.

So if we use h = 14.5 m, then:

\[
U = 5 × 9.8 × 14.5 = 710.5 \, \text{J}
\]

But total energy is only 709.5 J — so we can't have more PE than total energy.

So the only way is that the shot was not thrown straight up, so it has horizontal velocity at max height.

So at max height, KE is not zero.

Let’s assume that.

So:

Let’s define:

- At release: \( K_i = 562.5 \, \text{J}, U_i = 147 \, \text{J}, E = 709.5 \, \text{J} \)
- At max height: \( h = 14.5 \, \text{m} \), so \( U_f = mgh = 5 × 9.8 × 14.5 = 710.5 \, \text{J} \)

But 710.5 > 709.5 — impossible.

So conclusion: either the numbers are inconsistent, or we must use the given values as is.

But since the problem states both, perhaps it's a typo, or we should use the conservation principle.

Better approach: use the fact that energy is conserved, so at max height:

\[
U_{\text{max}} = E_{\text{total}} - K_{\text{max}}
\]

But we don't know \( K_{\text{max}} \)

But if it’s thrown straight up, then at max height, \( K = 0 \), and \( U = E_{\text{total}} = 709.5 \, \text{J} \)

Then height:

\[
h = \frac{U}{mg} = \frac{709.5}{5 × 9.8} = \frac{709.5}{49} = 14.48 \, \text{m}
\]

Which is approximately 14.5 m, so we can take it as such.

So we'll proceed with:

- At max height, PE = 709.5 J (since KE = 0)
- Height = 14.5 m (approximate)

But technically, if we use h = 14.5 m, then PE = 710.5 J, which is impossible.

So best to use conservation of energy and accept that the max height is slightly less than 14.5 m.

But the problem says "reaches a maximum height of 14.5 m", so we must use that.

So perhaps the initial speed is not 15.0 m/s, or mass is different.

Wait — maybe we should use the given max height to find the energy.

But the initial conditions are given.

Perhaps the shot is not thrown straight up — so at max height, it has horizontal velocity.

Let’s suppose that.

Let’s denote:
- At release: \( v = 15.0 \, \text{m/s} \), so \( K_i = 562.5 \, \text{J} \)
- \( U_i = 147 \, \text{J} \)
- Total energy: 709.5 J

At max height: \( h = 14.5 \, \text{m} \), so \( U_f = 5 × 9.8 × 14.5 = 710.5 \, \text{J} \)

But 710.5 > 709.5 — impossible.

So the only way this makes sense is if g is taken as 9.8, but the max height is not exactly 14.5 m, or the initial speed is not 15.0 m/s.

But the problem says both.

Perhaps there's a typo, and the max height is 14.4 m or 14.48 m.

Given the numbers, we should trust the conservation of energy.

So let's assume that at max height, the potential energy is equal to the total energy minus any remaining kinetic energy.

But if the shot is thrown at an angle, then at max height, it has horizontal velocity, so KE > 0.

But then PE < total energy.

But the problem asks for PE at max height.

So let’s calculate it as:

\[
U_{\text{max}} = mg h_{\text{max}} = 5.00 × 9.8 × 14.5 = 710.5 \, \text{J}
\]

But this exceeds total energy — impossible.

So the only logical resolution is that the maximum height is not 14.5 m, or the initial speed is not 15.0 m/s.

But since the problem states both, perhaps we should use the given values and see.

Wait — perhaps the initial height is 3.00 m, and max height is 14.5 m, so Δh = 11.5 m.

From energy conservation:

The increase in PE = mgΔh = 5 × 9.8 × 11.5 = 563.5 J

This comes from decrease in KE.

Initial KE = 562.5 J

But 563.5 > 562.5 — so not enough KE to gain that much PE.

So impossible.

So the only way is that the initial speed is higher.

But it's given as 15.0 m/s.

So conclusion: the numbers are inconsistent.

But in many textbooks, they round.

So perhaps they expect us to use:

- For (e): PE at max height = mg h_max = 5 × 9.8 × 14.5 = 710.5 J

Even though it exceeds total energy.

But that can't be.

Alternatively, perhaps the initial speed is not 15.0 m/s, but the vertical component is 15.0 m/s.

But the problem says "traveling at 15.0 m/s" — so total speed.

I think the intended solution is to use conservation of energy and accept that at max height, the potential energy is the total energy, and the height is approximately 14.5 m.

So let's do that.

So:

(e) Potential energy at max height:

Since at max height, if thrown straight up, KE = 0, so PE = total energy = 709.5 J

So answer: 709.5 J

Even though h = 14.5 m would give 710.5 J, we use the energy conservation.

So we'll go with that.

Similarly, for (f) and (g), at max height, KE = 0.

So:

(f) Kinetic energy at max height: 0 J

(g) Struck the ground:

When it hits the ground, h = 0, so PE = 0

Total energy = 709.5 J, so all is kinetic energy.

So KE = 709.5 J

Now let's write the answers.

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Final Answers for Problem 14:

(a) Potential energy at release:
\[
U = mgh = 5.00 \times 9.8 \times 3.00 = \boxed{147} \text{ J}
\]

(b) Kinetic energy at release:
\[
K = \frac{1}{2}mv^2 = \frac{1}{2} \times 5.00 \times (15.0)^2 = \boxed{562.5} \text{ J}
\]

(c) Total energy at release:
\[
E = 147 + 562.5 = \boxed{709.5} \text{ J}
\]

(d) Total energy at max height:
Conserved: \boxed{709.5} \text{ J}

(e) Potential energy at max height:
At max height, if thrown straight up, KE = 0, so PE = total energy = \boxed{709.5} \text{ J}

(f) Kinetic energy at max height:
\boxed{0} \text{ J}

(g) Kinetic energy when it strikes the ground:
All potential energy is converted to kinetic energy.
PE at release = 147 J, but total energy is 709.5 J, so at ground, KE = total energy = \boxed{709.5} \text{ J}

Note: The given max height of 14.5 m is slightly inconsistent with the initial data, but we resolve it using energy conservation.

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Would you like me to solve any other problems from the list?