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Chapter 13: Problem 20

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's speed is \(v_{A}=2.5 \mathrm{m} / \mathrm{s}\) directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3\) determine the speed at which each crate slides off the ramp at \(B .\) Assume that no tipping occurs. Take \(\theta=30^{\circ}.\)

Short Answer

Expert verified
Calculate the final speed by executing the steps: Firstly list the parameters, then calculate the work done by friction and finally apply energy conservation to find the final velocity. Substitute the equations and solve which gives us the answer.

Step by step solution

01

Identify Given Values

Firstly, let's list down the given parameters. The mass of crate \(m=12 \, kg\), initial speed \(v_{A}=2.5 \, m/s\), the coefficient of kinetic friction \(\mu_{k}=0.3\) and angle of the ramp \(\theta=30^{\circ}\).
02

Find work done by friction

The work done by friction is \(W_{f}=\mu_{k} * m * g * d\), where \(g\) is the acceleration due to gravity and \(d\) is the length of the slide, which is equal to the distance from A to B. Since there is no tipping and the angle is provided, we can use the trigonometric relationship to express d in terms of height h as \(d=h/\sin(\theta)\). And we can calculate \(h=v_{A}^2 / (2g)\) using the principle of conservation of energy. Substituting these two equations will give us \(W_{f}=\mu_{k} * m * g * v_{A}^2 / (2g * \sin(\theta))\).
03

Implement Law of Conservation of Mechanical Energy

According to the law of conservation of mechanical energy, the initial kinetic energy plus any work done equals the final kinetic energy. Therefore, we can solve for the final velocity \(V_{B}\) using the equation: \(1/2m*v_{A}^2 - W_{f} = 1/2m * v_{B}^2\). Reorganize the equation to solve for \(v_{B} = \sqrt{v_{A}^2 - 2 * W_{f} / m}\). Substitute in the equation from step 2 and solve for \(v_{B}\).

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is fundamental in physics and states that energy cannot be created or destroyed but only transformed from one form to another. When analyzing the motion of the crate on the ramp, we apply this principle to track the energy changes. At the top of the ramp (point A), all the energy is in the form of kinetic energy due to its speed. As the crate slides down, some of this kinetic energy gets converted into work done by friction, which is a form of energy lost to heat. The energy transformation can be summarized as:
  • Initial kinetic energy at A
  • Energy lost to friction as the crate slides down
  • Final kinetic energy at B
Understanding this energy conversion is crucial to calculate the final velocity of the crate at the bottom of the ramp. By knowing how much energy is "spent," we can deduce the remaining energy, which is transformed back into kinetic energy.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. For a moving object like the crate on the ramp, kinetic energy is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \]where \(m\) is the mass of the object, and \(v\) is its velocity. Initially, at point A, the crate has kinetic energy due to its given velocity of 2.5 m/s. As the crate slides down the ramp, its kinetic energy will decrease due to the opposing force of friction. Furthermore, the loss in this energy depends on the distance the crate travels and the force of friction acting over that distance. By correctly calculating the change in kinetic energy, we can predict the speed of the crate at the bottom of the ramp.
Work Done by Friction
When the crate moves along the ramp, kinetic friction acts opposite to the motion. The work done by friction is the energy used up to overcome this resistive force. It can be calculated using the formula:\[ W_f = \mu_k \times m \times g \times d \]where:
  • \( \mu_k \) is the coefficient of kinetic friction
  • \( m \) is the mass of the crate
  • \( g \) is the acceleration due to gravity
  • \( d \) is the distance the crate slides down the ramp
This formula highlights the factors influencing the work done by friction, making clear that more friction (a higher \( \mu_k \)) or a longer path \( d \) will increase the work done and decrease the crate's final velocity. By integrating the effects of friction into our energy calculations, we ensure a complete understanding of the motion dynamics and calculate a more realistic final speed at the bottom of the ramp.

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Most popular questions from this chapter

The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a lima莽on, \(r=(2+\cos \theta)\) ft. If \(\theta=\left(0.5 t^{2}\right)\) rad, where \(t\) is in seconds, determine the force which the rod exerts on the particle at the instant \(t=1\) s. The fork and path contact the particle on only one side.

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Block \(A\) has a mass \(m_{A}\) and is attached to a spring having a stiffness \(k\) and unstretched length \(l_{0} .\) If another block \(B,\) having a mass \(m_{B},\) is pressed against \(A\) so that the spring deforms a distance \(d\), determine the distance both blocks slide on the smooth surface before they begin to separate. What is their velocity at this instant?
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