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Chapter 7: Problem 47

A small box with mass \(0.600 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline that slopes upward at \(37.0^{\circ}\) above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is \(\mu_{k}=0.400\) The spring is released and the box travels up the incline, leaving the spring behind. What minimum elastic potential energy must be stored initially in the spring if the box is to travel \(2.00 \mathrm{~m}\) from its initial position to the top of the incline?

Short Answer

Expert verified
The calculation made in Step 2 provides the minimal potential energy stored in the spring. This implies that to cover the specified distance uphill overcoming friction and gravity, the spring must initially store this amount of potential energy.

Step by step solution

01

Identify the forces acting on the box

Firstly, two main forces are acting on the box as it moves up the incline, namely:1. The gravitational force: \( mg \cos(37.0^{\circ})\)2. The frictional force: \( \mu_{k} mg \cos(37.0^{\circ})\)
02

Calculate the work done against these forces

The work done against both the gravitational and the frictional forces can be calculated by the formula: \[ d = 2.00 m \]\[ W = d ( F_{g} + F_{f} ) \]To calculate for the work, input the given values into the equation:\[ W = 2.00 m \times ((0.600 kg \times 9.8 m/s^2 \times \cos(37.0^{\circ})) + (0.400 \times 0.600 kg \times 9.8 m/s^2 \times \cos(37.0^{\circ}))) \]
03

Calculate the elastic potential energy stored in the spring

The work done on the box by these forces is equal to the minimum elastic potential energy that must initially be stored in the spring for the box to be able to travel to the top of the incline. After performing the above calculation for work done, you have the answer for the stored potential energy.

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Key Concepts

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

Kinetic Friction
Imagine sliding a heavy box across the floor; the force you feel resisting the motion is kinetic friction. In our exercise, kinetic friction plays a crucial role as it opposes the movement of the box up the incline. It's the force that acts between moving surfaces and is calculated with the formula \( F_{k} = \mu_{k} \times N \), where \( \mu_{k} \) is the coefficient of kinetic friction and \( N \) is the normal force鈥攊n this case, \( mg \cos(\theta) \). Since the incline angle and mass are known, and gravity \( (g) \) is a constant \( 9.8 m/s^2 \), we can determine the force of kinetic friction accurately.

This force does work against the box, consuming some of the energy provided by the spring. We assess the energy lost to friction by multiplying the force of kinetic friction by the distance the box travels, demonstrating how essential it is to consider friction when calculating motion on surfaces.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics. It tells us that work done on an object is equal to the change in its energy. In the context of our exercise, the work done by the spring on the box is transformed into kinetic energy as the box begins to move, and some of this energy is then converted into elastic potential energy due to the box climbing the incline.

The work done against gravity and friction is given by the equation \( W = Fd \), where \( F \) is the total force working against the movement, and \( d \) is the distance the box travels. By calculating the work done against gravitational and frictional forces, we can determine the minimum elastic potential energy needed in the spring for the box to reach the specified distance on the incline. This illustrates how the work-energy principle links the stored energy in the spring to the mechanical work required to overcome opposing forces.
Gravitational Force
Every object on Earth experiences the pull of gravity, a force that draws objects toward the center of the Earth. The gravitational force on the box in our exercise is central to solving the problem. It is this force that the box must work against to travel up the incline.

Gravitational force is determined by the mass of the object and the acceleration due to gravity (\( g = 9.8 m/s^2 \) on Earth). On an incline, the effective gravitational force the box experiences is \( mg \cos(\theta) \) whereas \( \theta \) is the angle of the incline, and \( \cos(\theta) \) represents the fraction of the gravitational force acting parallel to the surface of the incline. Gravitational force contributes to the total work that needs to be done by the elastic potential energy stored in the spring, allowing us to understand how much energy is necessary for the box to ascend the given distance on the incline.

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Most popular questions from this chapter

A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m (Fig. E7.9). Assume that the size of the rock is small com-pared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J. (a) Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point B? (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

A small block with mass \(m\) slides without friction on the inside of a vertical circular track that has radius \(R .\) What minimum speed must the block have at the bottom of its path if it is not to fall off the track at the top of its path?

A conservative force \(\vec{F}\) is in the \(+x\) -direction and has magnitude \(F(x)=\alpha /\left(x+x_{0}\right)^{2},\) where \(\alpha=0.800 \mathrm{~N} \cdot \mathrm{m}^{2}\) and \(x_{0}=0.200 \mathrm{~m}\). (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 \(m=0.500 \mathrm{~kg}\) is released from rest at \(x=0\) and moves in the \(+x\) -direction. If \(\vec{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x=0.400 \mathrm{~m} ?\)

CALC A small block with mass \(0.0400 \mathrm{~kg}\) is moving in the \(x y\) -plane. The net force on the block is described by the potential-energy function \(U(x, y)=\left(5.80 \mathrm{~J} / \mathrm{m}^{2}\right) x^{2}-\left(3.60 \mathrm{~J} / \mathrm{m}^{3}\right) \mathrm{y}^{3}\). What are the magnitude and direction of the acceleration of the block when it is at the point \((x=0.300 \mathrm{~m}, y=0.600 \mathrm{~m}) ?\)

DATA You are designing a pendulum for a science museum. The pendulum is made by attaching a brass sphere with mass \(m\) to the lower end of a long. light metal wire of (unknown) length \(L\). A device near the top of the wire measures the tension in the wire and transmits that information to your laptop computer. When the wire is vertical and the sphere is at rest, the sphere's center is \(0.800 \mathrm{~m}\) above the floor and the tension in the wire is \(265 \mathrm{~N}\). Keeping the wire taut, you then pull the sphere to one side (using a ladder if necessary) and gently release it. You record the height \(h\) of the center of the sphere above the floor at the point where the sphere is released and the tension \(T\) in the wire as the sphere swings through its lowest point. You collect your results: \begin{tabular}{l|lllllll} \(h(\mathbf{m})\) & 0.800 & 2.00 & 4.00 & 6.00 & 8.00 & 10.0 & 12.0 \\ \hline \(\boldsymbol{T}(\mathrm{N})\) & 265 & 274 & 298 & 313 & 330 & 348 & 371 \end{tabular} Assume that the sphere can be treated as a point mass, ignore the mass of the wire, and assume that total mechanical energy is conserved through each measurement. (a) Plot \(T\) versus \(h,\) and use this graph to calculate \(L\). (b) If the breaking strength of the wire is \(822 \mathrm{~N}\), from what maximum height \(h\) can the sphere be released if the tension in the wire is not to exceed half the breaking strength? (c) The pendulum is swinging when you leave at the end of the day. You lock the museum doors, and no one enters the building until you return the next morning. You find that the sphere is hanging at rest. Using energy considerations, how can you explain this behavior?
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