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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 -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 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 force constant of the spring needed is approximately 7.91 N/m.

Step by step solution

01

Understanding the Problem

We need to calculate the force constant \( k \) of a spring so that a crate, after compressing the spring at the bottom of the ramp, does not rebound. Given:- The weight of each crate is 1470 N.- Initial velocity \( v_i = 1.8 \; \text{m/s} \).- Angle of the ramp \( \theta = 22.0^\circ \).- Friction force \( f_k = 550 \; \text{N} \).- Distance traveled is 8.0 m.
02

Calculating Initial Kinetic Energy

The initial kinetic energy \( KE_i \) of the crate is given by \( KE_i = \frac{1}{2} m v_i^2 \). First, calculate the mass \( m \) of the crate: \( m = \frac{1470}{9.8} \approx 150 \; \text{kg} \). Thus, \( KE_i = \frac{1}{2} \cdot 150 \cdot (1.8)^2 = 243 \; \text{J} \).
03

Calculating Work Done Against Friction

The work done by friction \( W_f \) is equal to the friction force times the distance it acts. Here, \( W_f = f_k \cdot d = 550 \cdot 8.0 = 4400 \; \text{J} \).
04

Calculating Change in Potential Energy

The change in potential energy \( \Delta PE \) as the crate moves down 8.0 meters along the ramp is given by \( \Delta PE = m g h \) where \( h = 8.0 \sin(22.0^\circ) \). So, \( \Delta PE = 150 \cdot 9.8 \cdot 8.0 \cdot \sin(22.0^\circ) \approx 4410 \; \text{J} \).
05

Applying Energy Conservation

According to energy conservation, the initial kinetic energy plus the potential energy change minus the work done against friction equals the energy stored in the spring, i.e.,\[ 243 + 4410 - 4400 = \frac{1}{2} k x^2 \].
06

Solving for the Spring Constant

The above equation simplifies to \[ 253 = \frac{1}{2} k (8.0)^2 \]. Solving for \( k \) gives:\[ k = \frac{2 \cdot 253}{64} \approx 7.91 \; \text{N/m} \].

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on two main factors: the mass of the object and its velocity. The formula for kinetic energy is given by:
  • \(KE = \frac{1}{2}mv^2\)
where \( m \) is the mass and \( v \) is the velocity. In the given exercise, the crate at the top of the ramp has an initial velocity, leading us to calculate its initial kinetic energy.

This energy is essential as it determines how much energy the crate has to spend as it moves down the ramp, facing frictional forces. Understanding kinetic energy helps us gauge the work performed by an object in motion, aiding the calculation of the required spring force constant.
Friction Force
Friction force is a resistive force that acts against the motion between two surfaces in contact. In this problem, a kinetic friction force opposes the movement of the crate as it slides down the ramp.

Kinetic friction can be calculated using the formula:
  • \(F_k = \mu_k N\)
However, the given kinetic friction force is directly provided as 550 N. This friction force does work on the moving crate, reducing its kinetic energy.
Recognizing the importance of friction helps us understand how much energy is dissipated as heat, which in turn affects the energy available for compressing the spring.
Conservation of Energy
The principle of conservation of energy states that the total energy in a system remains constant if it is isolated from external work or energy sources. This principle is crucial in analyzing the crate's journey down the ramp. In our scenario, the total initial energy (kinetic plus potential) is transformed due to work done by friction, resulting in the energy stored in the spring. The equation:
  • \(KE_i + \Delta PE - W_f = \frac{1}{2} k x^2\)
illustrates this transformation. The net energy, after accounting for losses due to friction, allows us to solve for the spring constant \( k \) necessary to ensure the crate stops without rebounding.
This principle underlines the fact that energy merely changes form rather than being created or destroyed.
Potential Energy
Potential energy is the energy stored in an object due to its position or configuration. For a crate on a slope, its potential energy depends on its vertical height and is given by:
  • \(PE = mgh\)
Here, \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is the height, which is part of the traveled distance. As the crate moves down the slope, this energy converts from potential to kinetic energy and other forms like thermal energy due to friction.
Incorporating potential energy into our calculations provides a more comprehensive picture of the energy transformations taking place as the crate slides downwards, ensuring the effective use of conservation of energy principle in our solution.

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Most popular questions from this chapter

The Great Sandini is a 60 -kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 \(\mathrm{N} / \mathrm{m} / \mathrm{m}\) that he will compress with a force of 4400 \(\mathrm{N}\) .The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40 \(\mathrm{N}\) during the 4.0 \(\mathrm{m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 \(\mathrm{m}\) above his initial rest position?

BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. \(A 2.0\) -mm-long, 0.50 -mg critter can reach a height of 20 \(\mathrm{cm}\) in a single leap. (a) Neglecting air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a \(65-\mathrm{kg}, 2.0-\mathrm{m}-\) tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would he need? (d) In fact, most humans can jump no more than 60 \(\mathrm{cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a 65-kg person? (e) Where does the flea store the energy that allows it to make such a sudden leap?

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0 -kg bucket 2.00 \(\mathrm{m}\) above the floor (Fig. \(\mathrm{P} 7.55 ) .\) Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. You can ignore friction and the mass of the pulley.

CALC A cutting tool under microprocessor control has several forces acting on it. One force is \(\vec{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{J}}\) , a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{3} .\) Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{m}\) .\(y=3.00 \mathrm{m} .\) (a) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) if this displacement is along the straight line \(y=x\) that connects these two points. (b) Calculate the work done on the tool by \(\vec{\boldsymbol{F}}\) if the tool is first moved out along the \(x\) -axis to the point \(x=3.00 \mathrm{m}, y=0\) and then moved parallel to the \(y\) -axis to the point \(x=3.00 \mathrm{m}, y=3.00 \mathrm{m}\) . (c) Compare the work done by \(\vec{\boldsymbol{F}}\) along these two paths. Is \(\vec{\boldsymbol{F}}\) conservative or nonconservative? Explain.

CALE A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) 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 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?
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