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Chapter 5: Problem 61

\(\bullet\) 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?

Short Answer

Expert verified
(a) The spring stretches 0.152 m; (b) it stretches 0.175 m.

Step by step solution

01

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 \( F = 9.50 \times 2.00 = 19.0 \text{ N} \).
02

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 \( x = \frac{19.0}{125} = 0.152 \text{ m} \).
03

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 \( F_s \cos(30.0^\circ) = 19.0 \text{ N} \).
04

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 \( x \), \( x = \frac{21.93}{125} = 0.175 \text{ m} \).

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Key Concepts

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

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/s虏. 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鈥檚 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, \( \theta = 30.0掳 \), thus: \( F_s \cos(30.0掳) = 19.0 \text{ N} \)
  • 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.

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Most popular questions from this chapter

\(\bullet\) An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope (Figure 5.38 ). The rope will break if the tension in it exceeds \(2.50 \times 10^{4} \mathrm{N} .\) Our hero's mass is 90.0 \(\mathrm{kg}\) . (a) If the angle \(\theta\) is \(10.0^{\circ}\) , find the tension in the rope. Start with a free-body diagram of the archaeologist. (b) What is the small- est value the angle \(\theta\) can have if the rope is not to break?

\(\bullet$$\bullet\) Prevention of hip injuries. People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. Let us investigate the worst-case sce- nario, in which a 55 kg person completely loses her footing (such as on icy pavement) and falls a distance of \(1.0 \mathrm{m},\) the dis- tance from her hip to the ground. We shall assume that the per- son's entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person's speed to 1.3 \(\mathrm{m} / \mathrm{s}\) over a distance of 2.0 \(\mathrm{cm} .\) Find the acceleration (assumed to be constant) of this person's hip while she is slow- ing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see if it is likely to cause injury, calcu- late how long it lasts.

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\(\bullet$$\bullet\) A man pushes on a piano of mass 180 \(\mathrm{kg}\) so that it slides at a constant velocity of 12.0 \(\mathrm{cm} / \mathrm{s}\) down a ramp that is inclined at \(11.0^{\circ}\) above the horizontal. No appreciable friction is acting on the piano. Calculate the magnitude and direction of this push (a) if the man pushes parallel to the incline, (b) if the man pushes the piano up the plane instead, also at 12.0 \(\mathrm{cm} / \mathrm{s}\) paral- lel to the incline, and (c) if the man pushes horizontally, but still with a speed of 12.0 \(\mathrm{cm} / \mathrm{s}\) .

\(\bullet$$\bullet\) Friction in an elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32\) what magnitude of force must you apply?
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