Problem 61 - College Physics

Question

\(\bullet\) 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; (b) it stretches 0.175 m.

Step 1: Determine Net Force

To find the force applied by the spring, we first need to calculate the net force acting on the sled using Newton's second law, \( F = ma \). Here, \( m = 9.50 \text{ kg} \) and \( a = 2.00 \text{ m/s}^2 \). So the net force is \( F = 9.50 \times 2.00 = 19.0 \text{ N} \).

Step 2: Calculate Spring Stretch in Horizontal Pull

For part (a), the spring pulls horizontally. The force from the spring is equal to the net force. Using Hooke's Law, \( F = kx \), where \( k = 125 \text{ N/m} \) is the spring constant and \( x \) is the stretch, we have \( 19.0 = 125x \). Solving for \( x \), we find \( x = \frac{19.0}{125} = 0.152 \text{ m} \).

Step 3: Apply Newton's Second Law at an Angle

For part (b), the spring pulls at an angle of \( 30.0^\circ \) above the horizontal. The force exerted by the spring needs to be decomposed into horizontal and vertical components. The horizontal component must still equal \( ma = 19.0 \text{ N} \). If \( F_s \) is the spring force, then \( F_s \cos(30.0^\circ) = 19.0 \text{ N} \).

Step 4: Solve for Spring Force and Stretch

From Step 3, \( F_s = \frac{19.0}{\cos(30.0^\circ)} \). Using \( \cos(30.0^\circ) = \frac{\sqrt{3}}{2} \), we find \( F_s = \frac{19.0}{\frac{\sqrt{3}}{2}} = \frac{38.0}{\sqrt{3}} \approx 21.93 \text{ N} \). Plug this value into Hooke's Law, \( 21.93 = 125x \). Solving for \( x \), \( x = \frac{21.93}{125} = 0.175 \text{ m} \).

Key Concepts

Newton's Second Law

Newton's Second Law of Motion is a fundamental principle in classical mechanics. It states that the force (\( F \)) acting on an object is equal to the mass (\( m \)) of that object multiplied by its acceleration (\( a \)). Mathematically, it's expressed as:
\[ F = ma \]
This law explains how the velocity of an object changes when it is subjected to an external force. In our exercise, the sled with a mass of 9.50 kg is accelerated at 2.00 m/s². Therefore, the net force required to produce this acceleration is calculated as:

• \( F = 9.50 \times 2.00 = 19.0 \text{ N} \)

This calculated force acts as the spring force necessary to pull the sled across the ice, demonstrating Newton's Second Law in action.

Hooke's Law

Hooke's Law describes the behavior of springs and how they stretch or compress. It states that the force (\( F \)) needed to extend or compress a spring by some distance (\( x \)) is proportional to that distance. The law can be written as:
\[ F = kx \]
where \( k \) is the spring constant, which reflects the stiffness of the spring. A higher \( k \) value means a stiffer spring. In our scenario, the spring constant is given as 125 N/m, indicating how much force is needed per meter of stretch.

• Horizontal stretch: Using \( F = kx \), we equate \( F \) with the net force from Newton’s second law (19.0 N) to find:
  • \( 19.0 = 125x \) implies \( x = \frac{19.0}{125} = 0.152 \text{ m} \)

Hooke's Law thus helps us understand the relationship between the force exerted by the spring and the amount it stretches or compresses.

Force Components

When a force acts at an angle, it does not apply its entire magnitude in one direction. Instead, it breaks down into two parts: horizontal and vertical components. Understanding these components is crucial for solving problems involving forces at an angle.

• Horizontal Component: To find the force acting purely in the horizontal direction, we use the cosine function:
  • For a force \( F_s \) acting at an angle \( \theta \), the horizontal component is \( F_s \cos(\theta) \).
  • In our exercise, \( \theta = 30.0° \), thus: \( F_s \cos(30.0°) = 19.0 \text{ N} \)
• Vertical Component: The vertical part of the force does not affect the acceleration along the sled's direction of motion but can affect normal forces or cause vertical displacement in other contexts.
  • It is given by \( F_s \sin(\theta) \), not needed in our specific exercise but useful in similar contexts.

These components help us deal with forces that don't act in straight lines, like the spring pulling the sled at an angle.

Spring Constant

The spring constant \( k \) is a measure of the stiffness of a spring. Higher values imply that more force is needed to stretch or compress the spring by a given amount, while lower values indicate a more easily stretched or compressed spring.

• In our practice problem, the spring constant is 125 N/m.
• It tells us that for every meter the spring is stretched or compressed, a force of 125 N is required.
• This is a key value when using Hooke's Law \( F = kx \).

With a known spring constant, we can solve for unknowns such as how much the spring stretches, given a certain amount of force. By applying this constant directly in our exercise, we manage to calculate the spring's extension under different conditions, whether it pulls horizontally or at an angle.