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Chapter 7: Problem 54

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

Short Answer

Expert verified
The spring constant needed is approximately 1650 N/m.

Step by step solution

01

Understanding the Problem

First, we need to clarify what is happening as the crate moves down the ramp. The crate, starting with an initial velocity of 1.8 m/s, will slide down a ramp and compress a spring. The spring must stop the crate without it rebounding. The forces acting on the crate include gravity, kinetic friction, and the spring force.
02

Energy Considerations

We use the principle of energy conservation. The initial kinetic energy plus potential energy (due to gravity) will be converted into the work done against friction and the spring potential energy. The initial kinetic energy is given by \[ KE_i = \frac{1}{2}mv^2 \] The potential energy due to gravity is \[ PE_g = mgh \] The work done against friction over the distance \(d\) is \[ W_f = F_{friction} \times d \] And the potential energy stored in the spring when compressed is \[ PE_s = \frac{1}{2}kx^2 \] where \(k\) is the spring constant and \(x\) is the compression.
03

Initial Energies

Calculate the initial kinetic energy using the mass derived from the weight:\[ m = \frac{1470 \text{ N}}{9.8 \text{ m/s}^2} \]Substitute into the kinetic energy formula:\[ KE_i = \frac{1}{2} \left( \frac{1470}{9.8} \right) (1.8)^2 \]
04

Potential Energy and Work Done by Friction

Determine the height \(h\) related to the ramp's slope:\[ h = 8.0 \times \sin(22.0^{\circ}) \]Calculate potenti... and work done by the friction force:\[ PE_g = \frac{1470}{9.8} \times 9.8 \times h = 1470\cdot h \]\[ W_f = 550 \times 8.0 \]
05

Balance Energy Equation

Set up the energy conservation equation:\[ \frac{1}{2}mv^2 + mhg = 550 \times 8 + \frac{1}{2}kx^2 \]This simplifies to:\[ \frac{1}{2} \left(\frac{1470}{9.8}\right)(1.8)^2 +1470\cdot 8 \cdot \sin(22) = 550\cdot 8 + \frac{1}{2}k8^2 \]
06

Solve for the Spring Constant \(k\)

Rearrange the equation to solve for \(k\):\[ \frac{1}{2}k\cdot8^2 = \left( \frac{1}{2} \left( \frac{1470}{9.8} \right)(1.8)^2 + 1470\cdot 8 \cdot \sin(22) - 550 \times 8 \right) \]\[ k = \frac{2\left( KE_i + PE_g - W_f \right)}{8^2} \]Calculate \(k\) by substituting all known values.

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Key Concepts

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

Kinetic Energy
Understanding kinetic energy is crucial in solving physics problems like the one involving the delivery ramp. Kinetic energy (\( KE \)) refers to the energy that an object possesses due to its motion. When a crate slides down the ramp, it converts potential energy into kinetic energy.The kinetic energy is calculated with the formula:\[ KE = \frac{1}{2} mv^2\]- \( m \) is the mass of the object and can be derived from the weight using the equation \( m = \frac{F_g}{g} \)- \( v \) is the velocity of the object, which in this case is the speed of the crate at the top of the ramp.In our problem, the crate starts with an initial velocity of 1.8 m/s. The initial kinetic energy is crucial because as the crate moves down the ramp, this kinetic energy is transferred into other forms of energy, eventually being used to compress the spring at the bottom.
Potential Energy
Potential energy is another vital concept when dealing with objects on an inclined plane. Potential energy (\( PE \)) is the stored energy of an object due to its position relative to a reference point, often the lowest point possible.For an object on a ramp, gravitational potential energy is given by:\[ PE = mgh\]- \( m \) is the mass,- \( g \) is the acceleration due to gravity (\( 9.8 \text{ m/s}^2 \)), and- \( h \) is the height of the ramp.The height can be determined by the angle of the ramp, as shown in the original exercise:\[ h = d \sin(\theta)\]Where \( d \) is the total distance along the ramp and \( \theta \) is the angle of inclination. This potential energy is converted into kinetic energy as the crate moves down, and then into the energy needed to compress the spring.
Spring Force
Spring force is a fundamental concept for understanding how and why the crate comes to rest when it compresses the spring at the bottom of the ramp. The force exerted by a spring is described by Hooke's Law, which says that the force needed to compress or extend a spring is proportional to the distance it is compressed or extended.The formula for spring force is:\[ F_s = kx\]- \( k \) is the spring constant, which measures the stiffness of the spring.- \( x \) is the compression distance.In this exercise, the spring's job is to absorb the kinetic energy of the crate and prevent it from rebounding. The energy stored in the spring as potential energy is given by:\[ PE_s = \frac{1}{2} kx^2\]By setting the potential energy in the spring equal to the net energy after accounting for friction, we can solve for \( k \). This calculation ensures that not only does the crate stop, but it also does not bounce back, achieving the desired design criteria.

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Most popular questions from this chapter

You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the middle of the gym floor, as shown in Fig. \(7.26 .\) You take your physics book and push it from one person to the other. The book has a mass of 1.5 \(\mathrm{kg}\) , and the coefficient of kinetic friction between the book and the floor is \(\mu_{\mathrm{x}}=0.25 .\) (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connecting these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by friction during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

A 10.0 -kg microwave oven is pushed 8.00 \(\mathrm{m}\) up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{F}\) with a magnitude 110 \(\mathrm{N}\) and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250 . (a) What is the work done on the oven by the force \(\vec{F} ?\) (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), \((b),\) and (c) to calculate the increase in the oven's kinetic energy. \((e)\) Use \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\) to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 8.00 \(\mathrm{m}\) . From this, compute the increase in the oven's kinetic energy, and compare it to the answer you got in part (d).

A \(2.00-\mathrm{kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) , compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. 7.30\()\) . (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta\) (a) Show that \(U(x)\) can be written as $$ U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$ Graph \(U(x)\) . Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x),\) the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x)\) . Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0} \text { and } x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

An object has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=\alpha x \hat{\imath},\) a force in the \(x\) -direction whose magnitude depends on the position of the object. (See Problem \(6.96 . )\) The constant is \(\alpha=2.00 \mathrm{N} / \mathrm{m}^{2} .\) The object moves along the following path: (1) It starts at the origin and moves along the \(y\) -axis to the point \(x=0\) , \(y=1.50 \mathrm{m} ;(2)\) it moves parallel to the \(x\) -axis to the point \(x=1.50 \mathrm{m}, y=1.50 \mathrm{m} ;(3)\) it moves parallel to the \(y\) -axis to the point \(x=1.50 \mathrm{m}, y=0 ;(4)\) it moves parallel to the \(x\) -axis back to the origin. (a) Sketch this path in the \(x y\) -plane. (b) Calculate the work done on the object by \(\overrightarrow{\boldsymbol{F}}\) for each leg of the path and for the complete round trip. (c) Is \(\overrightarrow{\boldsymbol{F}}\) conservative or nonconservative? Explain.
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