/*! 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 2 A 20-lb block slides down a \(30... [FREE SOLUTION] | 91Ó°ÊÓ

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A 20-lb block slides down a \(30^{\circ}\) inclined plane with an initial velocity of \(2 \mathrm{ft} / \mathrm{s}\). Determine the velocity of the block in \(3 \mathrm{~s}\) if the coefficient of kinetic friction between the block and the plane is \(\mu_{k}=0.25\).

Short Answer

Expert verified
The velocity of the block sliding down the plane after 3 seconds is approximately 29.45 ft/s.

Step by step solution

01

Calculate the Forces Acting on the Block

Firstly, separate the gravity force into two components: one perpendicular to the inclined plane (using \(\cos(30^{\circ})\)) and another one parallel to the plane (using \(\sin(30^{\circ})\)). The force parallel to the surface is \( F_{g\parallel} = W \sin (30^{\circ}) = 20 \times \sin(30^{\circ}) = 10 \, \mathrm{lb}\). The force perpendicular to the surface is \( F_{g\perp} = W \cos (30^{\circ}) = 20 \times \cos(30^{\circ}) \approx 17.32\, \mathrm{lb}\). The friction force ( \(F_f\) ), opposing the motion of the block, is equal to the kinetic friction coefficient times the normal force (the force with which the block is 'pushed' into the surface of the plane). Therefore, \(F_f = \mu_k \times F_{g\perp} = 0.25 \times 17.32\, \mathrm{lb} \approx 4.33\, \mathrm{lb}\).
02

Find the Net Force

The net force acting on the block along the inclined plane is the difference between the gravitational force parallel to the plane and the frictional force: \(F_{net} = F_{g\parallel} - F_f = 10\, \mathrm{lb} - 4.33\, \mathrm{lb} = 5.67\, \mathrm{lb}\).
03

Calculate the Acceleration

Next, find the acceleration of the block using Newton's second law: \( F = ma \). We know that 'm' is mass in slugs. We need to convert the weight (20lbs) into mass in slugs because we know that \( m = W / g \) where 'g' is the acceleration due to gravity which is 32.2 ft/s^2. Therefore, \( m = 20 / 32.2 \approx 0.62 \, \mathrm{slugs}\). Substitute \( F_{net} \) and 'm' into the equation and solve for 'a': \( a = F_{net} / m = 5.67 / 0.62 \approx 9.15 \, \mathrm{ft/s}^2 \).
04

Find the Final Velocity

The final velocity of the block after 3 seconds can now be found using the equation of linear motion: \( V_f = V_i + a t \), where \( V_i \) is the initial velocity, 'a' is acceleration and 't' is time. Substituting the known values, we get \( V_f = 2 + 9.15 \times 3 = 2 + 27.45 = 29.45 \, \mathrm{ft/s} \).

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Key Concepts

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

Understanding Kinetic Friction
Kinetic friction plays a central role when objects slide against each other. It is the force that opposes the relative motion between two surfaces in contact. Specifically, in an inclined plane problem, kinetic friction acts to resist the motion of an object sliding down the slope. The amount of kinetic friction can be calculated using the coefficient of kinetic friction, represented by \( \mu_k \), and the normal force, which is the force perpendicular to the surface of contact. In mathematical terms, the kinetic friction force is given by \( F_f = \mu_k \times F_{g\perp} \). Kinetic friction is crucial because it reduces the acceleration of the moving object. This introduces an important element in solving physics problems, as it must be accounted for to predict the motion accurately.
Applying Newton's Second Law
Newton's second law of motion is fundamental to understanding the dynamics of an object. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (\( F = ma \)). Applying this law to an inclined plane scenario, we determine the object's acceleration by dividing the net force along the incline by the object's mass. It's important to convert weight to mass before using this law, often requiring a conversion from pounds to slugs in U.S. customary units, using the relation \( m = W / g \), where \( g \) is the acceleration due to gravity. Newton's second law is the key framework for predicting how an object will move when forces are applied to it.
Breaking Down the Components of Forces
Dealing with forces on an inclined plane necessitates a clear understanding of the components of forces. The weight of an object can be resolved into two components on an inclined plane: a component perpendicular to the plane and a component parallel to the plane. The perpendicular component contributes to the normal force, which in turn affects kinetic friction. The parallel component drives the object down the slope and is given by \( W \sin(\theta) \) while the perpendicular component is \( W \cos(\theta) \). Understanding these components allows for the correct calculation of other forces, such as friction, and ultimately leads to the determination of the net force acting on the object.
Exploring Linear Motion on an Inclined Plane
Linear motion refers to the movement of an object along a straight path. In the context of an inclined plane, linear motion describes the movement of the object either up or down the slope. The final velocity of the object moving under constant acceleration, such as the gravitational force minus friction on an incline, can be determined using kinematic equations. The basic formula we use is \( V_f = V_i + at \), where \( V_f \) is the final velocity, \( V_i \) is the initial velocity, 'a' is constant acceleration, and 't' is the time elapsed. By applying this formula, you can find out how fast the object will be moving after a certain period. This is a key element of classical mechanics and enables prediction of future positions of moving objects.

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Most popular questions from this chapter

If the rod of negligible mass is subjected to a couple moment of \(M=\left(30 t^{2}\right) \mathrm{N} \cdot \mathrm{m}\), and the engine of the car supplies a traction force of \(F=(15 t) \mathrm{N}\) to the wheels, where \(t\) is in seconds, determine the speed of the car at the instant \(t=5 \mathrm{~s}\). The car starts from rest. The total mass of the car and rider is \(150 \mathrm{~kg} .\) Neglect the size of the car.

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Sand is deposited from a chute onto a conveyor belt which is moving at \(0.5 \mathrm{~m} / \mathrm{s}\). If the sand is assumed to fall vertically onto the belt at \(A\) at the rate of \(4 \mathrm{~kg} / \mathrm{s}\), determine the belt tension \(F_{B}\) to the right of \(A\). The belt is free to move over the conveyor rollers and its tension to the left of \(A\) is \(F_{C}=400 \mathrm{~N}\)

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