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Chapter 6: Problem 34

To stretch a spring 3.00 cm from its unstretched length, 12.0 J of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 cm from its unstretched length? (c) How much work must be done to compress this spring 4.00 cm from its unstretched length, and what force is needed to compress it this distance?

Short Answer

Expert verified
(a) \(26666.67 \, \text{N/m}\); (b) \(800 \, \text{N}\); (c) \(21.33 \, \text{J}\) and \(1066.67 \, \text{N}\).

Step by step solution

01

Understand Hooke's Law and Work Done

Hooke's Law relates the force exerted by a spring to its displacement. It is expressed as:\[ F = kx \]where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium. The work done on a spring is given by:\[ W = \frac{1}{2} k x^2 \]We use these formulas to solve the problems related to the spring.
02

Calculate the Force Constant (Spring Constant)

To find the spring constant \( k \), use the given work \( W = 12.0 \, \text{J} \) and displacement \( x = 3.00 \, \text{cm} = 0.03 \, \text{m} \). Substitute these into the work formula:\[ 12.0 = \frac{1}{2} k (0.03)^2 \]Solve for \( k \):\[ 12.0 = \frac{1}{2} k (0.0009) \]\[ 12.0 = 0.00045k \]\[ k = \frac{12.0}{0.00045} \approx 26666.67 \, \text{N/m} \]
03

Calculate Force to Stretch the Spring 3.00 cm

Using Hooke's Law, calculate the force required to stretch the spring 3.00 cm:\[ F = kx = 26666.67 \, \text{N/m} \times 0.03 \, \text{m} \]\[ F = 800 \, \text{N} \]
04

Calculate Work to Compress the Spring 4.00 cm

To find the work required to compress the spring 4.00 cm (\( x = 4.00 \, \text{cm} = 0.04 \, \text{m} \)), use the work formula:\[ W = \frac{1}{2} k x^2 \]\[ W = \frac{1}{2} \times 26666.67 \, \text{N/m} \times (0.04)^2 \]\[ W = \frac{1}{2} \times 26666.67 \times 0.0016 \]\[ W = 21.33 \, \text{J} \]
05

Calculate Force to Compress the Spring 4.00 cm

Using Hooke's Law, calculate the force needed:\[ F = kx = 26666.67 \, \text{N/m} \times 0.04 \, \text{m} \]\[ F = 1066.67 \, \text{N} \]

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Key Concepts

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

Spring Constant
Understanding the spring constant is like finding out how stiff a spring is. This is a crucial part of Hooke's Law. The spring constant, often denoted by the letter "k," tells us how much force is needed to stretch or compress a spring by a certain distance.
  • Hooke's Law says that the force needed to stretch or compress a spring is proportional to the distance it is stretched.
    The equation used here is: \( F = kx \)
  • The spring constant \( k \) tells you how stiff the spring is. A higher \( k \) means a stiffer spring.
To calculate \( k \), we use the formula for work done on a spring: \( W = \frac{1}{2} k x^2 \).
Solving for \( k \), given \( W = 12.0 \) J and \( x = 0.03 \) m, we substitute these into the equation and solve for \( k \) to find:
\( k \approx 26666.67 \) N/m. This result means that 26666.67 newtons of force are needed per meter to stretch or compress the spring.
Work Done
The concept of work done on a spring refers to the amount of energy needed to stretch or compress it.
  • The formula to calculate work done on a spring is: \( W = \frac{1}{2} k x^2 \).
  • This formula uses the spring constant \( k \) and the displacement \( x \) to determine the work \( W \).
When a spring is compressed or stretched, energy is stored in it as potential energy.
For a 3.00 cm stretch resulting in 12.0 J of work, the work done for a 4.00 cm compression is found by substituting \( x = 0.04 \) m into the work formula.
This calculation results in \( W = 21.33 \) J, showing more energy is required due to the additional distance compressed.
Force Exerted
The force exerted by a spring is a key part of understanding how springs work in everyday life. This force determines how hard you need to pull or push to stretch or compress the spring.
  • According to Hooke's Law: \( F = kx \).
  • Using this formula, we can calculate the force for any given displacement if we know \( k \).
To find the force needed to stretch the spring 3.00 cm, we use the previously calculated spring constant \( k = 26666.67 \) N/m and \( x = 0.03 \) m.
The force \( F \) calculated is 800 N.
Similarly, for a compression of 4.00 cm, the force required is found to be \( F = 1066.67 \) N.
These calculations show how the force increases with greater displacement, demonstrating the proportional relationship stated in Hooke's Law.

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Most popular questions from this chapter

A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

A soccer ball with mass 0.420 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?

When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

A surgeon is using material from a donated heart to repair a patient's damaged aorta and needs to know the elastic characteristics of this aortal material. Tests performed on a 16.0-cm strip of the donated aorta reveal that it stretches 3.75 cm when a 1.50-N pull is exerted on it. (a) What is the force constant of this strip of aortal material? (b) If the maximum distance it will be able to stretch when it replaces the aorta in the damaged heart is 1.14 cm, what is the greatest force it will be able to exert there?

A 5.00-kg package slides 2.80 m down a long ramp that is inclined at 24.0\(^\circ\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_k\) \(=\) 0.310. Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of the ramp, what is its speed after it has slid 2.80 m down the ramp?
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