Problem 76 - Physics

Question

An \(86.0-\mathrm{kg}\) climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{~m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Step-by-Step Solution

Answer

The rope stretches 1.026 m to break the climber's fall.

Step 1: Determine the gravitational force

To find the force acting on the climber due to gravity, we use the formula \( F = mg \), where \( m \) is the mass of the climber (86.0 kg), and \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)). Thus, the gravitational force is \( F = 86.0 \times 9.8 = 842.8 \text{ N} \).

Step 2: Calculate the potential energy at the point of fall

The potential energy (PE) of the climber right before the fall can be calculated using \( PE = mgh \) where \( h \) is the height fallen (0.750 m). Thus, \( PE = 86.0 \times 9.8 \times 0.750 = 631.5 \text{ J} \).

Step 3: Convert gravitational potential energy to elastic potential energy

When the climber falls and is momentarily at rest, all his initial gravitational potential energy is converted to the elastic potential energy stored in the rope. The formula for elastic potential energy is \( PE_{elastic} = \frac{1}{2}kx^2 \), where \( k \) is the spring constant (\(1.20 \times 10^{3} \text{ N/m}\)) and \( x \) is the displacement or stretch in the rope.

Step 4: Solve for the stretch in the rope

Using \( 631.5 = \frac{1}{2} \times 1.20 \times 10^3 \times x^2 \), solve for \( x \). Rearranging gives: \( x^2 = \frac{631.5 \times 2}{1.20 \times 10^3} \)Calculate \( x^2 = \frac{1263}{1200} = 1.0525 \). Taking the square root, \( x = \sqrt{1.0525} = 1.026 \text{ m} \).

Step 5: Confirm the solution

The stretch in the rope is confirmed by ensuring calculations match the conservation of energy principle where the gravitational potential energy is fully converted to elastic potential energy at the point when the climber comes to rest briefly. The calculated stretch \( x = 1.026 \text{ m} \) aligns with expectations from energy transformations.

Key Concepts

Gravitational Force

Gravitational force is one of the fundamental forces of nature and plays a critical role in many physics problems. It is the force that attracts any two masses towards each other. For objects near the Earth's surface, we often simplify this as the weight of the object.
The formula used to calculate gravitational force is:

• \( F = mg \)

where:

• \( F \) is the gravitational force in newtons (N)
• \( m \) is the mass of the object in kilograms (kg)
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \)

In our problem, this force was calculated for an 86 kg climber, resulting in a force of 842.8 N pulling him downward. This force is what causes the climber to fall after slipping, and it's the same force the safety rope must counteract.

Potential Energy

Potential energy is the stored energy of an object due to its position or state. In this context, we're focusing on gravitational potential energy, which depends on the object's height in a gravitational field.
The formula for gravitational potential energy is:

• \( PE = mgh \)

where:

• \( PE \) is the potential energy in joules (J)
• \( m \) is the mass in kilograms (kg)
• \( g \) is the gravity acceleration, 9.8 m/s²
• \( h \) is the height the object has fallen or moved in meters (m)

Before the climber begins to fall, he has gravitational potential energy due to his height above a reference point. As he falls, this energy is converted into kinetic energy and eventually into elastic potential energy as the rope stretches.

Elastic Potential Energy

Elastic potential energy is the energy stored in elastic materials as a result of their stretching or compressing. Think of it like storing energy in a spring. When an elastic object, like a rope, is stretched, it holds potential energy.
The formula to calculate elastic potential energy is:

• \( PE_{elastic} = \frac{1}{2}kx^2 \)

where:

• \( PE_{elastic} \) is the elastic potential energy in joules (J)
• \( k \) is the spring constant in newtons per meter (N/m)
• \( x \) is the displacement or stretch of the spring in meters (m)

In the climber's case, as they fall and the rope stretches, all gravitational potential energy is transformed into elastic potential energy. This stored energy ensures he comes to a stop, preventing any further fall.

Spring Constant

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress a spring by a unit distance.
A higher spring constant means a stiffer spring.

• The formula established by Hooke's law for the force \( F \) exerted by a spring is:

• \( F = kx \)

where:

• \( F \) is the force applied in newtons (N)
• \( k \) is the spring constant in newtons per meter (N/m)
• \( x \) is the distance the spring is stretched in meters (m)

For our exercise, the climber's rope has a spring constant of \( 1.20 \times 10^3 \text{ N/m} \), which influences how much the rope will stretch under the climber's weight and the gravitational force acting on it.

Energy Conservation

Energy conservation is a principle stating that energy cannot be created or destroyed, only transformed from one form to another. In physics problems, this concept helps in analyzing systems by ensuring all energy types add up to the same total.
For the falling climber, it's used to equate the initial potential energy with the energy stored in the stretched rope.

• The main principle here is:

• \( PE_{initial} = PE_{elastic} + KE_{other} \)

In this exercise, when the climber falls and comes to a stop, all of his gravitational potential energy is converted to elastic potential energy, showing there's no kinetic energy at that moment.
This transformation validates our calculations for the stretch in the rope and confirms the climber will remain safe temporarily halted by the rope's stretch.