/*! 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 207 The slider of mass \(m_{1}=0.4 \... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 207

The slider of mass \(m_{1}=0.4 \mathrm{kg}\) moves along the smooth support surface with velocity \(v_{1}=5 \mathrm{m} / \mathrm{s}\) when in the position shown. After negotiating the curved portion, it moves onto the inclined face of an initially stationary block of mass \(m_{2}=2 \mathrm{kg}\) The coefficient of kinetic friction between the slider and the block is \(\mu_{k}=0.30 .\) Determine the velocity \(v^{\prime}\) of the system after the slider has come to rest relative to the block. Neglect friction at the small wheels, and neglect any effects associated with the transition.

Short Answer

Expert verified
The velocity of the system after slider rests is approximately 0.833 m/s.

Step by step solution

01

Conservation of Momentum

Since the slider and the block form a system after the slider climbs onto the block, we need to apply conservation of momentum just before and just after the slider engages the block. Initially, the momentum is only from the slider since the block is stationary.\[ m_1 v_1 = (m_1 + m_2) v' \]Substitute the known values: \[ 0.4 \times 5 = (0.4 + 2) v' \]Simplifying gives us:\[ 2 = 2.4 v' \]
02

Solve for Initial Velocity

Solving for the velocity \( v' \), we get:\[ v' = \frac{2}{2.4} \approx 0.833 \text{ m/s} \]This is the velocity of the system immediately after the slider contacts the block.
03

Work Done by Friction

Now, consider the work done by friction to bring the slider to rest relative to the block. The work-energy principle will be used, where work done by friction equals the change in kinetic energy.The work done by friction \( W_f = \mu_k m_1 g \cdot d \), where \( d \) is the distance slider moves relative to the block.
04

Determine the Stopping Condition

Since the slider comes to rest relative to the block, the frictional force must remove all the relative kinetic energy.Thus, setting the work done equal to the initial kinetic energy relative to the block:\[ \frac{1}{2} m_1 v_{rel}^2 = \mu_k m_1 g \cdot d \]Substituting the known quantities:\[ \frac{1}{2} \times 0.4 \times (v_1 - v')^2 = 0.3 \times 0.4 \times 9.81 \times d \]
05

Solve for v_rel

Simplify and solve for \( v_{rel} = v_1 - v' \):\[ \frac{1}{2} \times 0.4 \times (5 - 0.833)^2 = 0.3 \times 0.4 \times 9.81 \times d \]Simplifying the left side, we find:\[ 4.166 = 1.1763 \cdot d \]
06

Solve for d

Re-arranging and solving for \( d \):\[ d = \frac{4.166}{1.1763} \approx 3.54 \text{ m}\]Since we found the kinetic energy entirely dissipated by friction, the slider has come to rest relative to the block.

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Key Concepts

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

Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding against each other. It plays a crucial role when the slider moves onto the block in this scenario.
  • The coefficient of kinetic friction, denoted by \( \mu_k \), is a measure of how slippery or rough a surface is.
  • Here, \( \mu_k \) is given as 0.30, meaning there is moderate friction between the slider and the block.
  • The force of kinetic friction \( f_k \) is calculated using: \( f_k = \mu_k \times N \), where \( N \) is the normal force. In our case, \( N = m_1 \times g \), the weight of the slider perpendicular to the surface.
Kinetic friction does work on the slider, converting its kinetic energy into thermal energy, thereby bringing it to rest relative to the moving block. This dissipation of energy is key to understanding how the slider's motion is slowed down and eventually stopped by the frictional force.
Work-Energy Principle
The work-energy principle states that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. In this problem, we focus on the work done by friction.
  • The initial kinetic energy of the slider when it contacts the block depends on its velocity just before touching the block, calculated from conservation of momentum.
  • The work done by the kinetic friction force \( W_f \) can be determined by \( W_f = \mu_k m_1 g \cdot d \), where \( d \) is the distance over which the frictional force acts.
  • To stop the slider relative to the block, the work done by friction must equal the initial relative kinetic energy of the slider.
  • This condition is set in the equation: \( \frac{1}{2} m_1 v_{rel}^2 = \mu_k m_1 g \cdot d \) and is used to solve for \( d \).
This principle illustrates how energy is transferred from kinetic energy to work done by friction, resulting in the slider eventually coming to stop relative to the block.
Relative Motion
Relative motion explains how the motion of the slider is observed in relation to the block it's moving upon.
  • Initially, the slider has a certain velocity \( v_1 \) with respect to the ground, while the block is stationary.
  • After contact, the slider and block form a system moving with a new velocity \( v' \).
  • To find the motion of the slider relative to the block, we calculate the velocity difference \( v_{rel} = v_1 - v' \).
  • This relative velocity decreases due to the kinetic friction acting between them until the slider comes to rest relative to the block.
By understanding relative motion, we see how the slider's speed changes from moving independently to a shared speed with the block, constantly influenced by the force of friction.

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Most popular questions from this chapter

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