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Ch 10: Interactions and Potential Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 10, Problem 57

CALC A 2.6 kg block is attached to a horizontal rope that exerts a variable force Fx = (20 − 5x) N, where x is in m. The coefficient of kinetic friction between the block and the floor is 0.25. Initially the block is at rest at x = 0 m. What is the block's speed when it has been pulled to x = 4.0 m?

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Step 1: Identify the forces acting on the block. The forces include the applied force \( F_x = 20 - 5x \) (in Newtons), the kinetic friction force \( F_k = \mu_k F_N \), where \( \mu_k = 0.25 \) is the coefficient of kinetic friction, and \( F_N \) is the normal force. Since the block is on a horizontal surface, \( F_N = mg \), where \( m = 2.6 \ \text{kg} \) and \( g = 9.8 \ \text{m/s}^2 \).
Step 2: Calculate the work done by the applied force \( F_x \) as the block moves from \( x = 0 \ \text{m} \) to \( x = 4.0 \ \text{m} \). The work done by a variable force is given by \( W_{\text{applied}} = \int_{x_1}^{x_2} F_x \, dx \). Substitute \( F_x = 20 - 5x \) and integrate from \( x = 0 \) to \( x = 4.0 \).
Step 3: Calculate the work done by the friction force. The friction force is constant and given by \( F_k = \mu_k mg \). The work done by friction is \( W_{\text{friction}} = -F_k \cdot d \), where \( d = 4.0 \ \text{m} \) is the displacement. The negative sign indicates that friction opposes the motion.
Step 4: Use the work-energy principle to find the block's speed. The work-energy principle states that the net work done on the block is equal to the change in its kinetic energy: \( W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}m(0)^2 \). Here, \( W_{\text{net}} = W_{\text{applied}} + W_{\text{friction}} \). Solve for \( v \), the final speed of the block.
Step 5: Substitute the values of \( W_{\text{applied}} \), \( W_{\text{friction}} \), \( m \), and simplify the equation to find \( v \). This will give the block's speed after it has been pulled to \( x = 4.0 \ \text{m} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This principle is crucial for analyzing the motion of the block, as it allows us to relate the applied force, friction, and the resulting acceleration to determine the block's speed.
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Work-Energy Principle

The Work-Energy Principle states that the work done on an object is equal to the change in its kinetic energy. In this scenario, we can calculate the work done by the variable force and the work done against friction to find the block's final kinetic energy and, consequently, its speed at x=4.0 m.
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Frictional Force

Frictional force opposes the motion of an object and is calculated as the product of the coefficient of kinetic friction and the normal force. In this problem, understanding the frictional force is essential, as it affects the net force acting on the block and thus influences its acceleration and final speed.
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Related Practice
Textbook Question

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