"""give me complete correct appropriate detailed answer in 4 steps

step 1 about introduction or overview of question

step 2 & 3 about answer with proper calculation and its formula , tables if r

Question

"""give me complete correct appropriate detailed answer in 4 steps

step 1 about introduction or overview of question

step 2 & 3 about answer with proper calculation and its formula , tables if required

also mention step titles
and last only final answer as per question asked format and requirements "


provide ALL THE ANSWER WITH FUIL CALVCULATION   WITHOUT AI VOICES STRICTLYA mountain climber is suspended from a nylon rope that has a force constant of 1.22 x 1044 N/m.

(2) What is the frequency (in Hz) at which the climber bounces, given that the combined mass of the climber
and his equipment is 85.0 kg?

(b) How much would the rope stretch (in cm) to break the climber's fall if he free-falls 2.00 m before the rope
runs out of slack? Hint: Use conservation of energy.

(0) Repeat both parts of the problem in a scenario where the length of the nylon rope is doubled.

VIPSolutions · Accepted Answer

**Step 1: Introduction / Overview of the Question**  
This problem involves a mountain climber suspended from a nylon rope. The rope acts as a spring with a given force constant (k = 1.22 × 10⁴ N/m). We need to find:  
(a) The frequency at which the climber bounces (simple harmonic motion).  
(b) The extension of the rope needed to stop the climber after a free-fall of 2.00 m, using energy conservation.  
(c) The effect of doubling the rope’s length on the above answers.

---

**Step 2: Frequency Calculation (Part a)**

**Formula:**  
The frequency (f) of oscillation for a mass-spring system is:  
$$ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} $$

**Given:**  
- Force constant, $ k = 1.22 \times 10^4 $ N/m  
- Mass, $ m = 85.0 $ kg

**Calculation:**  
$$
f = \frac{1}{2\pi} \sqrt{\frac{1.22 \times 10^4}{85.0}}
$$
$$
= \frac{1}{2\pi} \sqrt{143.53}
$$
$$
= \frac{1}{2\pi} \times 11.98
$$
$$
= \frac{11.98}{6.2832}
$$
$$
= 1.91 \text{ Hz}
$$

---

**Step 3: Rope Stretch Calculation Using Conservation of Energy (Part b)**

**Formula:**  
Potential energy lost by the climber = Elastic potential energy stored in the rope:  
$$
mgh = \frac{1}{2}kx^2
$$
Where:
- $ h = 2.00 $ m (fall distance)
- $ x $ = stretch in the rope

**Calculation:**
$$
(85.0)(9.8)(2.00) = \frac{1}{2}(1.22 \times 10^4)x^2
$$
$$
1666 = 6100x^2
$$
$$
x^2 = \frac{1666}{6100}
$$
$$
x^2 = 0.273
$$
$$
x = \sqrt{0.273}
$$
$$
x = 0.522 \text{ m}
$$

Convert to cm:
$$
x = 0.522 \times 100 = 52.2 \text{ cm}
$$

---

**Step 4: Repeat for Doubled Rope Length (Part c)**

**If rope length is doubled, the force constant (k) is halved:**
$$
k' = \frac{k}{2} = \frac{1.22 \times 10^4}{2} = 6.10 \times 10^3 \text{ N/m}
$$

**(a) New Frequency Calculation:**
$$
f' = \frac{1}{2\pi} \sqrt{\frac{k'}{m}}
$$
$$
= \frac{1}{2\pi} \sqrt{\frac{6.10 \times 10^3}{85.0}}
$$
$$
= \frac{1}{2\pi} \sqrt{71.76}
$$
$$
= \frac{1}{2\pi} \times 8.47
$$
$$
= \frac{8.47}{6.2832}
$$
$$
= 1.35 \text{ Hz}
$$

**(b) New Rope Stretch Calculation:**
$$
mgh = \frac{1}{2}k'x^2
$$
$$
1666 = 3050x^2
$$
$$
x^2 = \frac{1666}{3050}
$$
$$
x^2 = 0.546
$$
$$
x = \sqrt{0.546}
$$
$$
x = 0.739 \text{ m}
$$

Convert to cm:
$$
x = 0.739 \times 100 = 73.9 \text{ cm}
$$

---

**Final Answers:**

**(a) Frequency:** 1.91 Hz  
**(b) Rope stretch to stop fall:** 52.2 cm  
**(c) If rope length is doubled:**  
- New frequency: 1.35 Hz  
- New rope stretch: 73.9 cm

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