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The force constant of a spring is \(137 \mathrm{~N} / \mathrm{m}\). Find the magnitude of the force required to (a) compress the spring by \(4.80 \mathrm{~cm}\) from its unstretched length and (b) stretch the spring by \(7.36 \mathrm{~cm}\) from its unstretched length.

Short Answer

Expert verified
The magnitude of the force required to compress the spring by 4.80 cm from its unstretched length is 6.576 N. The magnitude of the force required to stretch the spring by 7.36 cm from its unstretched length is 10.0832 N.

Step by step solution

01

Understanding Hooke's Law

Hooke's law states that the force \(F\) needed to extend or compress a spring by some distance \(x\) is proportional to that displacement. This can be mathematically expressed as: \(F = kx\), where \(k\) is the spring constant.
02

Applying Hooke's Law (Part A)

Substitute the given values into Hooke's law: \(F = kx\). The spring constant \(k\) is given as \(137 \mathrm{~N} / \mathrm{m}\), and the displacement \(x\) is \(4.80 \mathrm{~cm}\) which we'll need to convert into meters for consistency. Now, \(F = 137 \times 0.048\).
03

Calculation (Part A)

Perform the multiplication to find the force: \(F = 6.576 \mathrm{~N}\). This is the magnitude of the force required to compress the spring by \(4.80 \mathrm{cm}\) from its unstretched length.
04

Applying Hooke's Law (Part B)

We will repeat step 2, this time for a displacement of \(7.36 \mathrm{~cm}\). Convert this displacement into meters and substitute into Hooke's law: \(F = 137 \times 0.0736\).
05

Calculation (Part B)

Perform the multiplication to find the force: \(F = 10.0832 \mathrm{~N}\). This is the magnitude of the force required to stretch the spring by \(7.36 \mathrm{~cm}\) from its unstretched length.

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Key Concepts

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

Spring Constant
When dealing with springs, understanding the spring constant is key. The spring constant, denoted as \( k \), measures a spring's stiffness. The stiffer the spring, the higher the value of \( k \). This means more force is required to compress or stretch the spring by a given distance. In physics, the spring constant is expressed in Newton per meter (N/m), representing the force needed per unit of length. For instance, if the spring constant \( k \) is 137 N/m, it signifies that 137 Newtons of force are needed to compress or extend the spring by 1 meter.
  • The spring constant is unique to each spring.
  • Its value gives an intuition about the force requirements.
  • Higher \( k \) values indicate stiffer springs.
Understanding this concept allows you to grasp how different springs react to various forces and helps apply Hooke's Law efficiently.
Force Calculation
Force calculation using Hooke's Law is quite straightforward. Hooke's Law is used to determine the force applied to a spring when it is stretched or compressed. The formula \( F = kx \) correlates the force \( F \) with the spring constant \( k \) and the displacement \( x \). First, ensure that the displacement is measured in meters to maintain the correct unit consistency. For example, our problem involves a spring constant \( k = 137 \) N/m, and displacements of 4.80 cm and 7.36 cm. These displacements convert into meters as 0.048 m and 0.0736 m, respectively.
  • The force is directly proportional to both displacement and spring constant.
  • Ensure all measurements are in compatible units before calculation.
By plugging these values into the formula, we calculate the necessary force to compress or stretch the spring. This methodology helps assess how much force different spring actions require.
Displacement in Meters
When working with Hooke's Law, emphasis on displacement measurement is crucial. Displacement refers to the change in length from the spring's natural, unstressed state. In most physics problems, it's essential to express displacement in meters. Doing so maintains unit consistency, making it compatible with the spring constant's units of Newton per meter (N/m). In the example provided, initially given in centimeters (cm), we convert these to meters by dividing by 100 since 1 meter = 100 centimeters. This simple step ensures the formula application is correct.
  • Always convert displacement to meters.
  • Accurate conversion is key for precise force calculations.
This method is an essential practice in physics, ensuring accurate calculations and successful solutions.

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Most popular questions from this chapter

A horizontal spring attached to a wall has a force constant of \(k=850 \mathrm{~N} / \mathrm{m}\). A block of mass \(m=1.00 \mathrm{~kg}\) is attached to the spring and rests on a frictionless, horizontal surface as in Figure P13.21. (a) The block is pulled to a position \(x_{i}=6.00 \mathrm{~cm}\) from equilibrium and released. Find the potential energy stored in the spring when the block is \(6.00 \mathrm{~cm}\) from equilibrium. (b) Find the speed of the block as it passes through the equilibrium position. (c) What is the speed of the block when it is at a position \(x_{i} / 2=3.00 \mathrm{~cm}\) ?

A child's toy consists of a piece of plastic attached to a spring (Fig. P18.11). The spring is compressed against the floor a distance of \(2.00 \mathrm{~cm}\), and the toy is released. If the toy has a mass of \(100 \mathrm{~g}\) and rises to a maximum height of \(60.0 \mathrm{~cm}\), estimate the force constant of the spring.

A student taking a quiz finds on a reference sheet the two equations $$ f=\frac{1}{T} \quad \text { and } \quad v=\sqrt{\frac{T}{\mu}} $$ She has forgotten what \(T\) represents in each equation. (a) Use dimensional analysis to determine the units required for \(T\) in each equation. (b) Explain how you can identify the physical quantity each \(T\) represents from the units.

A spring in a toy gun has a spring constant of \(9.80 \mathrm{~N} / \mathrm{m}\) and can be compressed \(20.0 \mathrm{~cm}\) beyond the equilibrium position. A \(1.00-\mathrm{g}\) pellet resting against the spring is propelled forward when the spring is released. (a) Find the muzzle speed of the pellet. (b) If the pellet is fired horizontally from a height of \(1.00 \mathrm{~m}\) above the floor, what is its range?

A \(0.250-\mathrm{kg}\) block resting on a frictionless, horizontal surface is attached to a spring having force constant \(\quad 83.8 \mathrm{~N} / \mathrm{m}\) as in Figure P13.16. A horizontal force \(\overrightarrow{\mathbf{F}}\) causes the spring to stretch a distance of \(5.46 \mathrm{~cm}\) from its equilibrium position. (a) Find the value of \(F\). (b) What is the total energy stored in the system when the spring is stretched? (c) Find the magnitude of the acceleration of the block immediately after the applied force is removed. (d) Find the speed of the block when it first reaches the equilibrium position. (e) If the surface is not frictionless but the block still reaches the equilibrium position, how would your answer to part (d) change? (f) What other information would you need to know to find the actual answer to part (d) in this case?

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