A 25.025.0-kg box of textbooks rests on a loading ramp that makes an angle αα with the horizontal. The coefficient of kinetic friction is 0.250.25, and the coefficient of static friction is 0.350.35. At this angle, find the acceleration once the box has begun to move.

Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 05: Applying Newton's Laws

Problem 36b

Chapter 5, Problem 36b

A $25.0$ -kg box of textbooks rests on a loading ramp that makes an angle $α$ with the horizontal. The coefficient of kinetic friction is $0.25$ , and the coefficient of static friction is $0.35$ . At this angle, find the acceleration once the box has begun to move.

Verified step by step guidance

Step 1: Begin by identifying the forces acting on the box. These include the gravitational force (weight), the normal force, and the force of kinetic friction. The gravitational force can be broken into two components: one parallel to the ramp ($ F_{g, \text{parallel}} = m g \sin \alpha $) and one perpendicular to the ramp ($ F_{g, \text{perpendicular}} = m g \cos \alpha $).

Step 2: Calculate the force of kinetic friction using the formula $ F_{\text{friction}} = \mu_k F_{\text{normal}} $, where $ \mu_k $ is the coefficient of kinetic friction and $ F_{\text{normal}} $ is the normal force. The normal force is equal to $ F_{g, \text{perpendicular}} $, which is $ m g \cos \alpha $.

Step 3: Determine the net force acting on the box along the ramp. The net force is the difference between the parallel component of the gravitational force and the force of kinetic friction: $ F_{\text{net}} = F_{g, \text{parallel}} - F_{\text{friction}} $. Substitute $ F_{g, \text{parallel}} = m g \sin \alpha $ and $ F_{\text{friction}} = \mu_k m g \cos \alpha $ into this equation.

Step 4: Use Newton's second law, $ F_{\text{net}} = m a $, to solve for the acceleration $ a $. Rearrange the equation to $ a = \frac{F_{\text{net}}}{m} $. Substitute $ F_{\text{net}} $ from Step 3 into this formula.

Step 5: Simplify the expression for acceleration: $ a = g (\sin \alpha - \mu_k \cos \alpha) $. This formula gives the acceleration of the box once it has begun to move. Plug in the given values for $ g $, $ \mu_k $, and $ \alpha $ to calculate the numerical result.

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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 essential for analyzing the forces acting on the box as it moves down the ramp, allowing us to calculate the acceleration by applying the formula F_net = m * a.

Frictional Forces

Frictional forces oppose the motion of an object and are categorized into static and kinetic friction. The coefficient of static friction applies when the box is at rest, while the coefficient of kinetic friction is relevant once the box starts moving. Understanding these coefficients helps determine the net force acting on the box as it accelerates down the ramp.

Inclined Plane Dynamics

When analyzing objects on an inclined plane, the gravitational force acting on the object can be resolved into components parallel and perpendicular to the surface. The angle of the ramp affects these components, influencing both the normal force and the frictional force, which are critical for calculating the net force and resulting acceleration of the box.

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Related Practice

Textbook Question

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are $0.355$ and $0.650$ , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to $30.0$ m/s without causing the box to slide? Draw a free-body diagram of the toolbox.

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Textbook Question

Two crates connected by a rope lie on a horizontal surface (Fig. E $5.37$ ). Crate A has mass $m_A$ , and crate B has mass $m_B$ . The coefficient of kinetic friction between each crate and the surface is $μ_k$ . The crates are pulled to the right at constant velocity by a horizontal force $F$ . Draw one or more free-body diagrams to calculate the following in terms of $m_A$ , $m_B$ , and $μ_k$ : the tension in the rope connecting the blocks.

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Textbook Question

Two crates connected by a rope lie on a horizontal surface (Fig. E $5.37$ ). Crate A has mass $m_{A}$ , and crate B has mass $m_{B}$ . The coefficient of kinetic friction between each crate and the surface is $μ_{k}$ . The crates are pulled to the right at constant velocity by a horizontal force $F$ . Draw one or more free-body diagrams to calculate the following in terms of $m_{A}$ , $m_{B}$ , and $μ_{k}$ : the magnitude of $F$ .

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Textbook Question

A $45.0$ -kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds $313$ N. Then you must reduce your push to $208$ N to keep it moving at a steady $25.0$ cm/s. Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is $1.62$ m/s².

(i) What magnitude push would cause it to move?

(ii) What would its acceleration be if you maintained the push in part (b)? Note: Part (b) asked what push you must exert to give it an acceleration of $1.10$ m/s².

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Textbook Question

A $45.0$ -kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds $313$ N. Then you must reduce your push to $208$ N to keep it moving at a steady $25.0$ cm/s. What push must you exert to give it an acceleration of $1.10$ m/s²?

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Textbook Question

You throw a baseball straight upward. The drag force is proportional to $v^{2}$ . In terms of $g$ , what is the $y$ -component of the ball's acceleration when the ball's speed is half its terminal speed and it is moving up?

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