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Chapter 7: Problem 48

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掳, the force component can be calculated using trigonometry:
  • Gravitational force component along the ramp: \( F = W \sin(\theta) \)
  • Calculation: \( W_{g} = 1470 \sin(22.0掳) \times 5 = 2760 \, \text{J} \)
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 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?

A bungee cord is 30.0 m long and, when stretched a distance \(x\), it exerts a restoring force of magnitude \(kx\). Your father-in-law (mass 95.0 kg) stands on a platform 45.0 m above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 m before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 380.0 N. When you do this, what distance will the bungee cord that you should select have stretched?

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 baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m/s and directed at an angle of 53.1\(^\circ\) above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1\(^\circ\) \(below\) the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?

While a roofer is working on a roof that slants at 36\(^\circ\) above the horizontal, he accidentally nudges his 85.0-N toolbox, causing it to start sliding downward from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?
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