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A certain spring found not to obey Hooke's law exerts a restoring force \(Fx(x) = -ax - \beta x^2\) if it is stretched or compressed, where \(\alpha\) = 60.0 N/m and \(\beta\) = 18.0 N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(\(x\)) for this spring. Let \(U = 0\) when \(x = 0\). (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the \(+x\)-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the \(x = 0\) equilibrium position?

Short Answer

Expert verified
The potential energy function is \( U(x) = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 \). The object's speed at 0.50 m when \( x = 0 \) is 7.00 m/s.

Step by step solution

01

Understand the force function

The restoring force given is \( F(x) = -\alpha x - \beta x^2 \) with \( \alpha = 60.0 \) N/m and \( \beta = 18.0 \) N/m\(^2\). This force is not linear due to the \( -\beta x^2 \) term.
02

Derive the potential energy function

The potential-energy function \( U(x) \) is found by integrating the force function. Since force is the negative gradient of potential energy, \( F(x) = -\frac{dU}{dx} \). Hence, \( \frac{dU}{dx} = \alpha x + \beta x^2 \). Integrate this with respect to \( x \) to get the potential energy function:\[ U(x) = \int (\alpha x + \beta x^2) \, dx = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 + C \]Given \( U(0) = 0 \), solve for \( C \). \( C = 0 \), so the potential energy function is:\[ U(x) = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 \]
03

Identify initial conditions for kinetic energy analysis

An object with mass 0.900 kg is attached to the spring and stretched to \( x = 1.00 \) m. Its initial potential energy is \( U(1.00) \), and initially, the speed is zero, so kinetic energy is zero.
04

Calculate initial potential energy

Substitute \( x = 1.00 \) m into the potential energy function:\[ U(1.00) = \frac{1}{2} (60.0) (1.00)^2 + \frac{1}{3} (18.0) (1.00)^3 = 30.0 + 6.0 = 36.0 \, \text{Joules} \]
05

Calculate mechanical energy conservation

Since the system is frictionless, mechanical energy conservation applies:\[ U_i + K_i = U_f + K_f \] The initial kinetic energy \( K_i = 0 \), initial potential energy \( U_i = 36.0 \) J, final potential energy \( U_f \), and final kinetic energy \( K_f \). We're solving for the speed of the object when \( x = 0.50 \) m.
06

Calculate potential energy at x = 0.50 m

Substitute \( x = 0.50 \) m into \( U(x) \):\[ U(0.50) = \frac{1}{2} (60.0) (0.50)^2 + \frac{1}{3} (18.0) (0.50)^3 = 7.5 + 0.75 = 8.25 \, \text{Joules} \]
07

Solve for final kinetic energy and speed

Using energy conservation, \( U_i - U_f = K_f \):\[ 36.0 - 8.25 = K_f = 27.75 \, \text{Joules} \]The final kinetic energy \( K_f = \frac{1}{2} m v^2 \). Solve for \( v \):\[ v = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(27.75)}{0.900}} \approx 7.00 \, \text{m/s} \]

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Key Concepts

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

Restoring Force
When we discuss the concept of a **restoring force**, we are basically looking at how a spring brings back an object to its equilibrium position. When a spring is compressed or stretched, it exerts a force attempting to return the object to its original state. Usually, springs obey Hooke's Law, where the force is linear with respect to displacement. This means the force is directly proportional to how much the spring is stretched or compressed. However, some springs, like the one in our problem, do not follow this simple rule. Instead, they exhibit a non-linear force behavior. The force given by such a spring is described by the formula:\[ F(x) = -\alpha x - \beta x^2 \]Here, the force is not a straight line but a curve, due to the inclusion of the term \( -\beta x^2 \). This means that as the displacement increases, not only does the force grow stronger, but it changes with the square of the distance too. This non-Hookean behavior implies complexities in how the spring acts on the object over larger distances, adding quadratic influence to the linear restore of the spring.
Potential Energy
**Potential energy** in the context of springs is the energy stored in the spring when it is stretched or compressed. For a non-Hookean spring, the potential energy is not a straightforward calculation. Yet, we can find it by integrating the restoring force over the distance.For the force function given:\[ F(x) = -\alpha x - \beta x^2 \]We integrate to find the potential energy function, \( U(x) \):\[ U(x) = \int (\alpha x + \beta x^2) \, dx = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 + C \]The constant \( C \) is determined by the condition that the potential energy is zero when the spring is at its unstretched position \((x = 0)\), giving us:\[ U(x) = \frac{1}{2} \alpha x^2 + \frac{1}{3} \beta x^3 \]This accounts for the total stored energy in the spring depending on how far it has been displaced. Understanding how this energy is stored and can be transformed is crucial for solving more complex motion problems.
Mechanical Energy Conservation
One important principle in physics is the **conservation of mechanical energy**. It states that in a closed system without external forces like friction, the total mechanical energy (sum of kinetic and potential energy) remains constant.For our spring system on a frictionless surface:\[ U_i + K_i = U_f + K_f \]Where \( U_i \) and \( U_f \) are the initial and final potential energies, and \( K_i \) and \( K_f \) are the initial and final kinetic energies. In our example, the spring is initially stretched, giving it potential energy, but no kinetic energy until it begins to move. As it returns towards equilibrium and reaches 0.50 m from the origin, some potential energy converts to kinetic energy:- Initial potential energy at 1.00 m is calculated as 36.0 Joules.- Potential energy at 0.50 m is 8.25 Joules.When these energies reassess at 0.50 m from balance, the remainder transforms into kinetic energy:\[ K_f = 36.0 - 8.25 = 27.75 \text{ Joules} \]The formula for kinetic energy \( K = \frac{1}{2} mv^2 \) helps us solve for the object's speed, showing that the energy conservation principle offers a direct path to understanding motion in such systems.

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Most popular questions from this chapter

In one day, a 75kg mountain climber ascends from the 1500m level on a vertical cliff to the top at 2400 m. The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 m. What is her change in gravitational potential energy (a) on the first day and (b) on the second day?

A 1.20-kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k =\) 1800 N/m that is compressed 15.0 cm. When the spring is released, how high does the cheese rise from this initial position? (The cheese and the springare \(not\) attached.)

A conservative force \(\overrightarrow{F}\) is in the \(+x\)-direction and has magnitude \(F(x) = a/(x + x_0)^2\), where \(\alpha = 0.800\) N \(\cdot\) m\(^2\) and \(x_0 = 0.200\) m. (a) What is the potential-energy function \(U(x)\) for this force? Let \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\). (b) An object with mass \(m = 0.500\) kg is released from rest at \(x = 0\) and moves in the \(+x\)-direction. If \(\overrightarrow{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x = 0.400\) m?

A spring of negligible mass has force constant \(k =\) 800 N/m. (a) How far must the spring be compressed for 1.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a 1.60-kg book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

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