/*! 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 12 A \(30.0\) -kg block is resting ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(30.0\) -kg block is resting on a flat horizontal table. On top of this block is resting a \(15.0\) kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is \(325 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the lower block and the table is \(0.600\), and the coefficient of static friction between the two blocks is \(0.900\). A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of the force \(\overrightarrow{\mathbf{F}}\).

Short Answer

Expert verified
(a) Spring compression is approximately 0.407 m. (b) Force magnitude is approximately 396.9 N.

Step by step solution

01

Determine Force of Static Friction

Calculate the maximum static friction force that can act on the upper block before it begins to slip. The force of static friction is given by the formula: \( f_s = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. Here, \( N = m_2 \cdot g = 15.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \). Calculate \( f_s = 0.900 \times 15.0 \times 9.8 = 132.3 \, \text{N} \).
02

Spring Compression Calculation

When the upper block is on the verge of slipping, the force of static friction equals the restoring force of the spring. The spring force \( f_s = k \cdot x \), gives us the equation: \( 132.3 = 325 \cdot x \). Solve for \( x \) to find the compression of the spring: \( x = \frac{132.3}{325} \approx 0.407 \, \text{m} \).
03

Determine Kinetic Friction Force

Calculate the kinetic friction force that opposes the motion of the lower block on the table using: \( f_k = \mu_k \cdot N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N = (m_1 + m_2) \cdot g = (30.0 + 15.0) \times 9.8 \). Thus, \( N = 45.0 \times 9.8 = 441 \, \text{N} \). Compute \( f_k = 0.600 \times 441 = 264.6 \, \text{N} \).
04

Calculate Force F

The net force acting on the 30 kg block is the applied force \( F \) minus the kinetic friction force \( f_k \). To keep the blocks moving at a constant speed, \( F - f_k = f_s \). Substitute \( f_s \) and \( f_k \) into the equation to find \( F \): \( F - 264.6 = 132.3 \). Solve for \( F \): \( F = 396.9 \, \text{N} \).

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Key Concepts

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

Spring Constant
The spring constant, often denoted as "\(k\)," is a fundamental characteristic of a spring. It quantifies the stiffness of a spring. The higher the spring constant, the stiffer the spring, meaning it requires more force to compress or stretch the spring by a certain amount. Mathematically, the spring constant is represented in the formula for Hooke's Law:
  • \(F_s = k \cdot x\)
Where:
  • \(F_s\) is the force exerted by the spring, measured in Newtons (N)
  • \(k\) is the spring constant, measured in Newtons per meter (N/m)
  • \(x\) is the displacement or compression of the spring from its equilibrium position, measured in meters (m)
In the given exercise, the spring constant is 325 N/m, which means it requires 325 Newtons of force to compress the spring by 1 meter. This value is crucial as it directly correlates with how much force is required to achieve a certain compression, as seen in the calculation.Understanding the spring constant helps in various applications, from mechanical devices to measuring how much energy a spring can store.
Kinetic Friction
Kinetic friction is the force resisting the movement of two bodies sliding against each other. It plays a significant role in understanding motion, especially in scenarios involving sliding blocks or cars on the road. Unlike static friction, which acts on objects that are not moving, kinetic friction only applies to objects in motion.The force of kinetic friction \(f_k\) is calculated using the formula:
  • \(f_k = \mu_k \cdot N\)
Where:
  • \(\mu_k\) is the coefficient of kinetic friction, a dimensionless value representing the ratio between the force of friction and the normal force.
  • \(N\) is the normal force, often equivalent to the weight of the object (\(m \cdot g\), where \(m\) is mass and \(g\) is the acceleration due to gravity).
In the problem, the coefficient of kinetic friction \(\mu_k\) is 0.600 for the lower block sliding on the table, and the normal force is calculated based on the masses of both blocks. This force opposes the direction of motion and is essential for determining the net force required to keep the blocks moving at a constant speed.Knowing how to calculate and understand kinetic friction is crucial for solving real-world problems involving motion.
Compression of Spring
The compression of a spring refers to the displacement of the spring from its natural length, typically due to an applied force. This concept is vital in the study of mechanical systems where springs are used for storing potential energy and for absorption of shock.In the given exercise, the compression of the spring \(x\) is related to the maximum static friction force that acts just before slipping occurs.To find the compression, the equation according to Hooke's Law is used:
  • \(f_s = k \cdot x\)
Where \(f_s\) is the static friction force found in the problem to be 132.3 N, and \(k\) is the spring constant.Hence, solving for \(x\) gives us:
  • \(x = \frac{f_s}{k} = \frac{132.3}{325} \approx 0.407 \, \text{m}\)
This calculation of 0.407 meters tells us how much the spring is compressed from its resting position when the upper block reaches the verge of slipping.Understanding spring compression is critical in designing systems that require resistance to force or changes in their potential energy state.

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Most popular questions from this chapter

An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{~N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is \(480 \mathrm{~N} / \mathrm{m}\). What is the displacement of the bowstring?

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{~m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as that of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{~m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

The front spring of a car's suspension system has a spring constant of \(1.50 \times 10^{6} \mathrm{~N} / \mathrm{m}\) and supports a mass of \(215 \mathrm{~kg}\). The wheel has a radius of \(0.400 \mathrm{~m}\). The car is traveling on a bumpy road, on which the distance between the bumps is equal to the circumference of the wheel. Due to resonance, the wheel starts to vibrate strongly when the car is traveling at a certain minimum linear speed. What is this speed?

A rifle fires a \(2.10 \times 10^{-2}\) kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{~m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{~m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

A person who weighs \(670 \mathrm{~N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{~cm}\). (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{~cm}\) ?

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