/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 106 A box with mass \(m\) sits at th... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 106

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

Short Answer

Expert verified
The coefficient of kinetic friction between the box and the ramp is given by \(\mu_k = (3 v_{0}^{2}) / (8 g d cos(\alpha)\). Note that this is independent of the box's mass, as expected for the friction coefficient.

Step by step solution

01

Find Initial and Final Kinetic Energy

By the principle of energy conservation, we can say that the initial kinetic energy of the box is \(E_{0} = 1/2 m v_{0}^{2}\). Also, the final kinetic kinetic energy at its starting position will be \(E_{f} = 1/2 m (v_{0}/2)^{2}= 1/8 m v_{0}^{2}\)
02

Find Work Done by Force of Friction

The work done by the force of friction equals the change in kinetic energy of the box. As total work done by the friction is negative (as it opposes motion), we can find it as follows: \(W_{fric} = E_{f} - E_{0} = 1/8 m v_{0}^{2} - 1/2 m v_{0}^{2} = -(3/8) m v_{0}^{2}\)
03

Identify Work Done by Friction and Normal Force

We can also write work done by friction in another way: \(W_{fric} = -f_k d\), where \(f_k\) is kinetic friction force. Also, friction force is \(f_k = \mu_k N\), where \(\mu_k\) is the kinetic friction coefficient and \(N\) is the normal force. Due to incline at angle \(\alpha\), normal force differs from box mass: \(N = m g cos(\alpha)\)
04

Calculate the Kinetic Friction Coefficient

Now we can compare expressions for the work done by the force of friction, to find the kinetic friction coefficient. Solving \(W_{fric} = -f_k d = -\mu_k N d = -(3/8) m v_{0}^{2}\) for \(\mu_k\) yields \(\mu_k = (3 v_{0}^{2}) / (8 g d cos(\alpha)\)

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental concept in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, we apply energy conservation to understand the movement of the box.

The box initially possesses kinetic energy due to its speed at the bottom of the ramp:
  • The initial kinetic energy, (\[E_0 = \frac{1}{2} m v_0^2\] ), is the energy derived from its initial velocity \(v_0\). This represents how the motion of the box holds energy.
  • When the box reaches back its starting point, its kinetic energy has decreased to (\[E_f = \frac{1}{8} m v_0^2\] ) as the speed becomes \(v_0/2\).
Understanding energy conservation helps us explain the changes in motion as the box travels upward and returns downward on the ramp, showing shifts from kinetic energy to other forms of energy like work done by friction.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In the context of this problem, it plays a crucial role in understanding how the box moves up and down the ramp.
  • Initially, the box moves with an energy based on its velocity \(v_0\), which is calculated using the formula (\[E_0 = \frac{1}{2} m v_0^2\]). This energy is what allows the box to overcome the gravitational pull and move up the ramp.
  • As the box escalates and reaches its highest point, it loses some kinetic energy due to friction work and the component of gravitational force acting against its direction of motion.
  • When it begins to slide back down, it again experiences kinetic energy. However, the downward journey is influenced by the energy lost to friction, resulting in only \(\frac{1}{8} m v_0^2\) kinetic energy when it gets back to the start.
Kinetic energy calculations reveal how much work friction does by comparing energy at different points of motion.
Work Done by Friction
Work done by friction is a dissipative force, meaning it removes energy from the system by converting kinetic energy into other forms like heat. In our scenario, friction works to slow down the box as it travels. Understanding this force provides insight into the coefficient of kinetic friction.

The work done by friction \(W_{fric}\) indicates how energy is taken out of the system:
  • The decrease in kinetic energy from \(1/2 m v_{0}^{2}\) to \(1/8 m v_{0}^{2}\) equals the work done by friction: \[W_{fric} = -(3/8) m v_{0}^{2}\].
  • This negative work shows energy loss, as friction opposes the movement of the box.
  • It can also be calculated by substituting friction force into the work equation: \[W_{fric} = -\mu_k N d = -(3/8) m v_{0}^{2}\]. Comparing these helps find \(\mu_k\).
    Here, \(\mu_k\), the coefficient of kinetic friction, measures how much friction force opposes motion relative to the normal force, \(N = m g \cos(\alpha)\). This enables the calculation: \[\mu_k = \frac{3 v_{0}^{2}}{8 g d \cos(\alpha)}\].
Understandably, friction and its work component are crucial in real-world applications, determining how quickly or easily an object decelerates upon contact with a surface.

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Most popular questions from this chapter

Block \(B\) has mass \(5.00 \mathrm{~kg}\) and sits at rest on a horizontal, frictionless surface. Block \(A\) has mass \(2.00 \mathrm{~kg}\) and sits at rest on top of block \(B\). The coefficient of static friction between the two blocks is \(0.400 .\) A horizontal force \(\vec{P}\) is then applied to block \(A .\) What is the largest value \(P\) can have and the blocks move together with equal accelerations?

A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is \(0.400 .\) A force \(\vec{F}\) is applied to the crate in a direction \(30.0^{\circ}\) above the horizontal. The minimum value of \(F\) required to get the crate to start sliding is \(380 \mathrm{~N}\). What is the mass of the crate?

A large crate with mass \(m\) rests on a horizontal floor. The coefficients of friction between the crate and the floor are \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{k}} .\) A woman pushes downward with a force \(\overrightarrow{\boldsymbol{F}}\) on the crate at an angle \(\theta\) below the horizontal. (a) What magnitude of force \(\vec{F}\) is required to keep the crate moving at constant velocity? (b) If \(\mu_{\mathrm{s}}\) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of \(\mu_{\mathrm{s}}\)

The Trendelenburg Position. After emergencies with major blood loss, a patient is placed in the Trendelenburg position, in which the foot of the bed is raised to get maximum blood flow to the brain. If the coefficient of static friction between a typical patient and the bed sheets is \(1.20,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.
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