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(III) An elevator cable breaks when a 925 -kg elevator is 22.5\(\mathrm { m }\) above the top of a huge spring \(( k =\) \(8.00 \times 10 ^ { 4 } \mathrm { N } / \mathrm { m } )\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; \(( b )\) the speed of the elevator just betore striking the spring; \(( c )\) the amount the spring compresses (note that here work is done by both the spring and gravity.

Short Answer

Expert verified
Work done by gravity is 203,452.125 J; speed is 21 m/s; spring compresses 0.258 m.

Step by step solution

01

Calculate Work Done by Gravity

The work done by gravity can be calculated using the formula: \[ W_g = mgh \] where \( m = 925 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( h = 22.5 \text{ m} \). Substitute the values to find \( W_g \): \[ W_g = 925 \times 9.81 \times 22.5 \approx 203,452.125 \text{ J} \]
02

Calculate Speed of the Elevator Before Impact

Use the work-energy principle which states that the work done by gravity equals the change in kinetic energy: \[ W_g = \frac{1}{2}mv^2 \] Solving for \( v \): \[ v = \sqrt{\frac{2W_g}{m}} \] Substitute the values: \[ v = \sqrt{\frac{2 \times 203,452.125}{925}} \approx 21 \text{ m/s} \]
03

Calculate Spring Compression

Set up the energy conservation equation between gravitational potential energy, kinetic energy, and elastic potential energy to solve for the compression \( x \): \[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 + mgx \] Substitute known values to find \( x \): \[ 203,452.125 = 40,000x^2 + 9,074.25x \] Solving this quadratic equation gives us \( x \approx 0.258 \text{ m} \)

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Key Concepts

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

Gravitational Potential Energy
Gravitational Potential Energy is an important form of energy associated with the position of an object in a gravitational field. Imagine an elevator hanging high above the ground. The further it is from the earth, the higher its gravitational potential energy. This is because gravity can do more work on it as it falls. This kind of energy depends on three main factors:

  • The mass of the object (\( m \))
  • The height of the object above a reference point (\( h \))
  • The gravitational acceleration (\( g \)), typically \( 9.81 \, \text{m/s}^2 \) on Earth.

Mathematically, gravitational potential energy (\( U_g \)) is calculated by the formula:
\[ U_g = mgh \]In our scenario, as the elevator was 22.5 meters above the spring, its potential energy was due to gravity pulling it downwards. This potential energy converts into kinetic energy as the elevator descends, influencing how fast it falls.
Kinetic Energy
Kinetic Energy is the energy an object possesses due to its motion. As the elevator moves downward, the potential energy it initially had converts into kinetic energy, making the elevator move faster. The formula for calculating kinetic energy (\( KE \)) is:
\[ KE = \frac{1}{2}mv^2 \]
Where:
  • \( m \) = mass of the object
  • \( v \) = velocity of the object
To find out how fast the elevator was moving just before it impacted the spring, we used the work-energy principle, which notes that the work done by forces such as gravity manifests as changes in kinetic energy. In this scenario, the entire work done by gravity during the elevator's descent transforms into the kinetic energy of the elevator right before it hits the spring.
This acceleration was calculated using the expressions for work and kinetic energy to find a velocity of about 21 m/s right before impact.
Spring Compression
Spring Compression involves the transfer of energy when an object compresses a spring. In this case, when the elevator hits the spring, its kinetic energy is transferred to the spring and potential energy due to the spring's compression. This is described by Hooke's Law, which tells us how springs behave. The compression (\( x \)) introduces elastic potential energy described by:
\[ E_s = \frac{1}{2}kx^2 \]
Where:
  • \( k \) = spring constant (a measure of the spring's stiffness)
  • \( x \) = amount of spring compression

To find out how much the spring compressed, we apply the principle of conservation of energy. The elevator's kinetic energy before impact equaled the sum of the spring's potential energy and the gravitational work done during the compression. Solving this balance gives a compression distance of approximately 0.258 meters. Energy conservation is crucial: it illustrates how the energies convert and assures us that energy isn't lost, merely shifted between forms in physical systems.

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Most popular questions from this chapter

A train is moving along a track with constant speed \(v_{1}\) relative to the ground. A person on the train holds a ball of mass \(m\) and throws it toward the front of the train with a speed \(v_{2}\) relative to the train. Calculate the change in kinetic energy of the ball \((a)\) in the Earth frame of reference, and (b) in the train frame of reference. (c) Relative to each frame of reference, how much work was done on the ball? (d) Explain why the results in part \((b)\) are not the same for the two frames-after all, it's the same ball.

One car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by \(7.0 \mathrm{~m} / \mathrm{s}\), they then have the same kinetic energy. What were the original speeds of the two cars?

(I) A 265 -kg load is lifted 23.0 m vertically with an acceler- ation \(a = 0.150 g\) by a single cable. Determine \(( a )\) the tension in the cable; \(( b )\) the net work done on the load; assuming it started from rest.

A 2200-N crate rests on the floor. How much work is required to move it at constant speed (a) \(4.0 \mathrm{~m}\) along the floor against a drag force of \(230 \mathrm{~N}\), and (b) \(4.0 \mathrm{~m}\) vertically?

When different masses are suspended from a spring, the spring stretches by different amounts as shown in the Table below. Masses are ±1.0 gram. \begin{tabular}{lllrrrrrrr} \hline Mass (g) & 0 & 50 & 100 & 150 & 200 & 250 & 300 & 350 & 400 \\ Stretch (cm) & 0 & 5.0 & 9.8 & 14.8 & 19.4 & 24.5 & 29.6 & 34.1 & 39.2 \\ \hline \end{tabular} (a) Graph the applied force (in Newtons) versus the stretch (in meters) of the spring, and determine the best-fit straight line. (b) Determine the spring constant ( \(\mathrm{N} / \mathrm{m}\) ) of the spring from the slope of the best-fit line. \((c)\) If the spring is stretched by \(20.0 \mathrm{~cm},\) estimate the force acting on the spring using the best-fit line.

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