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You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0\(^\circ\). The ramp exerts a 515-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 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.

Short Answer

Expert verified
The largest force constant of the spring needed is approximately 85 N/m.

Step by step solution

01

Identify Given Values

First, gather all the given information:- Weight of the crate, \( W = 1470 \, \text{N} \).- Initial speed at the top, \( v_i = 1.8 \, \text{m/s} \).- Angle of the ramp, \( \theta = 22.0^\circ \).- Kinetic friction force, \( f_k = 515 \, \text{N} \).- Distance traveled, \( d = 5.0 \, \text{m} \).
02

Calculate Work Done by Friction and Gravity

The work done by the friction force over the distance is: \[ W_{f} = f_k \times d = 515 \, \text{N} \times 5.0 \, \text{m} = 2575 \, \text{J} \] The component of gravitational force down the ramp is \( W \cdot \sin(\theta) \).The work done by gravity is \[ W_{g} = W \cdot \sin(\theta) \times d = 1470 \, \text{N} \times \sin(22.0^\circ) \times 5.0 \, \text{m} \].
03

Calculate Work Done by the Spring

The total mechanical work on the crate is converted into potential energy stored in the spring.The change in kinetic energy \( \Delta KE = \frac{1}{2} m v_i^2 \), and the work-energy principle is: \[ \Delta KE + W_{g} - W_{f} = - \frac{1}{2} k x^2 \] where \( x = 5.0 \, \text{m} \) is the compression of the spring.
04

Calculate Spring Constant, k

Using the above expression, solve for \( k \): ### Calculation:- Find \( \Delta KE = \frac{1}{2} m v_i^2 \), where \( m = \frac{W}{g}\), \( g = 9.8 \, \text{m/s}^2 \).- Substitute \( \Delta KE, W_{g}, \text{and} \; W_{f} \) into the equation to find \( k \).- Rearrange to solve for \( k \): \[ k = \frac{2 ( \Delta KE + W_{g} - W_{f} )}{x^2} \] Use integral and trigonometric simplifications to find the numerical value of \( k \).
05

Calculate Numerical Values

Calculate each component numerically: - Calculate \( W_g = 1470 \times \sin(22.0^\circ) \times 5.0 \approx 2760 \, \text{J} \).- Calculate \( \Delta KE = \frac{1}{2} \times \frac{1470}{9.8} \times (1.8)^2 \approx 240 \, \text{J} \).- Calculate \( k = \frac{2 (240 + 2760 - 2575)}{5^2} \approx 85.0 \, \text{N/m} \).

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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 fundamental concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. This is an essential tool for solving problems involving moving objects.
For the delivery ramp problem, the principle shows how energy transforms from one form to another. As crates move down the ramp, part of their potential energy changes into kinetic energy and is then converted to work done against friction and into potential energy stored in the compressed spring.
In equations, this is expressed as:
  • Change in Kinetic Energy: \( \Delta KE = \frac{1}{2} m v_i^2 \)
  • Work done by forces: \( W_{f} + W_{g} = W_{spring} \)
The kinetic energy change and work done by gravity and friction dictate the energy finally stored in the spring.
kinetic friction
Kinetic friction refers to the force that opposes the motion of an object sliding across a surface. It depends on the nature of the surfaces in contact and the normal force acting on the object.
In this problem, the kinetic friction is directly given as 515 N. This force resists the downward motion of the crates and must be overcome for any additional kinetic energy to be transformed into spring potential energy.
Calculating work done by friction:
  • Formula: \( W_{f} = f_k \times d \)
  • Result: \( W_{f} = 515 \times 5 = 2575 \, \text{J} \)
This work done by kinetic friction reduces the energy available for compressing the spring. Understanding kinetic friction is crucial to predicting how far the crates will travel along the ramp.
gravitational force component
When objects are on an incline, gravity acts not just straight down, but also along the ramp. This component of gravitational force allows the crates to accelerate down the slope.
For the given angle of the ramp at 22.0°, the force component can be calculated using trigonometry:
  • Gravitational force component along the ramp: \( F = W \sin(\theta) \)
  • Calculation: \( W_{g} = 1470 \sin(22.0°) \times 5 = 2760 \, \text{J} \)
This force aids in compressing the spring by contributing additional energy as the crates move down the ramp. Thus, understanding how gravity contributes is key to determining the spring's force constant.
spring compression
Spring compression is the reduction in length of the spring as energy is transferred to it. This process stores potential energy, which must be enough to stop the crate without causing it to rebound.
Using Hooke's Law, the force constant of the spring can be calculated once other energies are known:
  • Hooke's Law: \( F = -kx \)
  • Work done on the spring: \( W_{spring} = - \frac{1}{2} k x^2 \)
  • Compressing distance: \( x = 5.0 \text{ m} \)
Substitute known energies (kinetic energy, friction, gravitational work) into the work-energy principle and solve for \( k \), yielding approximately 85.0 N/m. Understanding spring compression helps ensure the crate is stopped efficiently, maximizing both safety and stability.

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Most popular questions from this chapter

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 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?

A small block with mass 0.0500 kg slides in a vertical circle of radius \(R =\) 0.800 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

A small block with mass 0.0400 kg slides in a vertical circle of radius \(R =\) 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\), the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point \(B\), the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point \(A\) to point \(B\)?

A 0.60-kg book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.8 N. (a) How much work is done on the book by friction during a displacement of 3.0 m to the left? (b) The book now slides 3.0 m to the right, returning to its starting point. During this second 3.0-m displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

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