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Cable Breaks The cable of the 1800 \(\mathrm{kg}\) elevator cab in Fig. \(10-54\) snaps when the cab is at rest at the first floor, where the cab bottom is a distance \(d=3.7 \mathrm{~m}\) above a cushioning spring whose spring constant is \(k=0.15 \mathrm{MN} / \mathrm{m}\). A safety device clamps the cab against guide rails so that a constant frictional force of \(4.4 \mathrm{kN}\) opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance \(x\) that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

Step by step solution

01

- Convert Units

Convert the given data into suitable units: Mass of the elevator cab: \(m = 1800 \mathrm{~kg}\)Spring constant: \(k = 0.15 \mathrm{MN/m} = 0.15 \times 10^6 \mathrm{N/m} = 150000 \mathrm{~N/m}\)Frictional force: \(f = 4.4 \mathrm{kN} = 4.4 \times 10^3 \mathrm{N} = 4400 \mathrm{~N}\)distance from cab bottom to spring: \(d=3.7 \mathrm{~m}\)

02

- Find Speed of the Cab before Hitting the Spring

Use the work-energy principle. The gravitational potential energy lost by the cab converts into kinetic energy and work done against friction.Potential energy lost: \(U = mgh = 1800 \,\mathrm{kg} \times 9.8 \,\mathrm{m/s^2} \times 3.7 \,\mathrm{m} = 65160 \,\mathrm{J}\)Work done against friction: \(W_f = f \times d = 4400 \,\mathrm{N} \times 3.7 \,\mathrm{m} = 16280 \,\mathrm{J}\)Total energy available for kinetic energy: \(65160 \,\mathrm{J} - 16280 \,\mathrm{J} = 48880 \,\mathrm{J}\)Using kinetic energy formula: \(KE = \frac{1}{2}mv^2 = 48880 \,\mathrm{J}\)Solving for \(v\): \(v = \sqrt{\frac{2 \times 48880}{1800}} \approx 5.22 \,\mathrm{m/s}\)

03

- Find Maximum Compression of the Spring

When the cab contacts the spring, its kinetic energy plus the work against friction will convert into the potential energy stored in the spring.Friction continues to do work against the cab motion, so let the total distance the spring compresses be \(x\).The work done against friction force during compression: \(W_f = f \times x = 4400 \,\mathrm{N} \times x\)Energy equation:\(\frac{1}{2} mv^2 + mgx = \frac{1}{2}kx^2 + f \times x\)Substitute values and solve numerically:\(\frac{1}{2} \times 1800 \times 5.22^2 + 1800 \times 9.8 \times x = \frac{1}{2} \times 150000x^2 + 4400x\)\(24564 + 17640x = 75000x^2 + 4400x\)Rearrange to form a quadratic equation:\(75000x^2 - 13240x - 24564 = 0\)Solve for \(x\): \(x \approx 0.614 \,\mathrm{m}\)

04

- Bouncing Back Distance

The cab will bounce back up due to the energy stored in the compressed spring minus the work done by friction. The energy equation when spring releases can be simplified to; since friction continues to act, from conservation of energy:\(\frac{1}{2}kx_{max}^2 - f \times x_{max} = mgx'\)Substitute \(x_{max} = 0.614 \,\mathrm{m}\): \(\frac{1}{2} \times 150000 \times 0.614^2 - 4400 \times 0.614 = 1800 \times 9.8 \times x'\)\(28126.65 - 2701.6 = 17640x'\)Solve for \(x'\): \(x' \approx 1.44\)

05

- Total Distance Moved

Find the approximate total distance the cab will move before coming to rest using the energy approach and including friction work continuously:Total distance calculation involves repeated motion until energy is exhausted due to work against friction. Hence, we can approximate considering the net energies used and emphasizing cumulative calculations.However, we approximate the sequence: 1st Compression: Friction dissipated: 3.7m drop & 0.614 compression.1st Rebound: Rise: 1.44m due friction energy subtraction again.Sequence converges as frictions continually exhausting energy.Estimating distance cycles duly diminishing for final rest-state beyond initial only approximating larger parts.Thus, accounting: Initial travel plus major initial rebounds sum into significant distances.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work-energy principle

The work-energy principle is a core concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. More formally, this can be expressed as:\[ W = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i \]
where \( W \) is the work done, \( m \) is the mass of the object, and \( v_f \) and \( v_i \) are the final and initial velocities, respectively.

In our exercise, when the elevator cab falls and hits the spring, the gravitational potential energy lost by the cab gets converted into kinetic energy and work done against the friction force. This is why the speed just before hitting the spring can be calculated using the work-energy principle.

Spring constant

The spring constant is a measure of a spring's stiffness. It's represented by \( k \) in equations and has units of force per unit length (e.g., Newtons per meter).

The fundamental equation for a spring is given by Hooke's Law:
\[ F = kx \]
Here, \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position.

In this scenario, the spring constant of the cushioning spring below the elevator cab is \( 0.15 \, \text{MN/m} \), which we converted to \( 150,000 \, \text{N/m} \) for easier calculations.

Gravitational potential energy

Gravitational potential energy (GPE) is the energy an object possesses because of its position in a gravitational field. It can be calculated using:
\[ U = mgh \]
where \( U \) is the gravitational potential energy, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (typically \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height above the reference point.

In our example, when evaluating the energy transformations of the cab, we started by calculating its gravitational potential energy at the height of \( 3.7 \, \text{m} \) above the spring.

Kinetic energy conversion

Kinetic energy (KE) is the energy possessed by an object due to its motion. The kinetic energy is given by:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass, and \( v \) is the velocity of the object.

In the exercise, when the elevator cab falls, its gravitational potential energy converts into kinetic energy. The frictional force acting on the cab reduces the total energy available for this conversion, as calculated in the solution.

Frictional force

Frictional force is the force that opposes the relative motion of two surfaces in contact. It's calculated as:
\[ F_f = \text{Coefficient of friction} \times \text{Normal force} \]
In our exercise, the frictional force acting on the elevator cab is a given constant: \( 4.4 \, \text{kN} \) or \( 4400 \, \text{N} \).

This force does work against the motion of the cab, which impacts the cab's speed, spring compression, and the distance the cab bounces back.

Quadratic equation in physics

Many problems in physics result in quadratic equations that we need to solve. The general form is:
\[ ax^2 + bx + c = 0 \]
We use the quadratic formula to find solutions:
\[ x = \frac{-b \, \text{±} \, \text{√}(b^2 - 4ac)}{2a} \]
In this exercise, when we calculated the maximum compression \( x \) of the spring, we formulated a quadratic equation by equating the kinetic energy and work against friction to the potential energy stored in the spring.