Problem 51 - College Physics

Question

A light spring having a force constant of \(125 \mathrm{~N} / \mathrm{m}\) is used to pull a \(9.50 \mathrm{~kg}\) sled on a horizontal frictionless ice rink. If the sled has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2},\) by how much does the spring stretch if it pulls on the sled (a) horizontally, (b) at \(30.0^{\circ}\) above the horizontal?

Step-by-Step Solution

Answer

(a) The spring stretches 0.152 m horizontally. (b) The spring stretches 0.175 m at 30°.

Step 1: Determine the Horizontal Force Required

Given that the sled has a mass of \(9.50 \mathrm{~kg}\) and an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\), use Newton's Second Law \(F = ma\) to find the force required to accelerate the sled. \[F = 9.50 \times 2.00 = 19.00 \mathrm{~N}\]

Step 2: Calculate the Stretch of the Spring (Horizontally)

The spring force is given by Hooke's Law \( F = kx \), where \(k\) is the spring constant and \(x\) is the stretch of the spring. Solve for \(x\) when the spring pulls horizontally:\[19.00 = 125x \quad \Rightarrow \quad x = \frac{19.00}{125} = 0.152 \mathrm{~m}\]

Step 3: Determine the Effective Force at an Angle

When the spring pulls at a \(30.0^{\circ}\) angle, only the horizontal component of the force contributes to the acceleration. This component is given by \(F_{x} = F \cos(\theta)\). Therefore:\[F_{x} = 19.00 \div \cos(30.0^{\circ}) \approx 19.00 \div 0.866 = 21.94 \mathrm{~N}\]

Step 4: Calculate the Stretch of the Spring (At an Angle)

Use Hooke's Law again to find the stretch of the spring when pulling at an angle:\[21.94 = 125x \quad \Rightarrow \quad x = \frac{21.94}{125} = 0.175 \mathrm{~m}\]

Key Concepts

Hooke's Law

Hooke's Law is a principle that relates the force exerted by a spring to its displacement from the equilibrium position. It is expressed through the formula \( F = kx \). Here, \( F \) is the force applied by the spring, \( k \) is the force constant of the spring, and \( x \) represents the stretch or compression of the spring. Hooke's Law assumes that the spring remains within its elastic limit, where it can still return to its original shape without permanent deformation. This law is fundamental when analyzing systems using springs because it provides a clear relationship between force and displacement, helping us predict how far a spring will stretch or compress under a given force. Mainly, it's crucial for determining how the performance of a spring changes with varying conditions, such as different orientations or external forces.

Force Constant

The force constant, represented by \( k \), is a value indicating the stiffness of a spring. Measured in Newtons per meter (N/m), it dictates how much force is needed to stretch or compress the spring by a unit distance. A higher force constant means a stiffer spring, which requires more force to stretch it. In the exercise, the spring has a force constant of \( 125 \, \mathrm{N/m} \). Understanding the force constant is essential because it allows you to calculate the stretch of the spring when a specific force is applied. Knowing \( k \) and using Hooke’s Law, you can easily find how a spring behaves under loading conditions. It helps in designing mechanical systems where specific spring behaviors are needed.

Spring Stretch

Spring stretch refers to how much a spring elongates when a force is applied. Using Hooke's Law, you can determine the stretch \( x \) by rearranging the formula to \( x = \frac{F}{k} \). In the horizontal scenario of the exercise, a force of 19.00 N results in a stretch of the spring by \( 0.152 \, \mathrm{m} \). When force changes due to additional factors like angles, the stretch changes as well, requiring recalculation using the adjusted force value. My understanding is knowing precisely how much a spring stretches is important in practical applications where exact distances and forces impact the overall system performance, such as in the suspension systems of vehicles or measuring forces and distances in controlled laboratory experiments.

Horizontal Force

Horizontal force is a component of the total force acting in the horizontal plane. In the simplest scenario, this force directly affects motion with no angle of deviation, as seen in the exercise. Newton’s Second Law \( F = ma \) helps calculate the horizontal force necessary to move the sled, accounting for only the mass and acceleration of the object.
In real-world applications, it is crucial to consider horizontal forces when designing and analyzing systems subject to various loads and forces that could affect their motion. Understanding horizontal forces allows engineers to predict how objects will behave under applied loads and helps in optimizing system designs to ensure safety and effectiveness. This becomes significantly more useful when dealing with non-horizontal forces, where determining the horizontal component is crucial for analysis and understanding of real-world scenarios like inclined planes or when additional forces act at angles.