Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 436

Natural length of a spring is \(60 \mathrm{~cm}\), and its spring constant is \(2000 \mathrm{~N} / \mathrm{m}\). A mass of \(20 \mathrm{~kg}\) is hung from it. The extension produced in the spring is..... $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4.9 \mathrm{~cm}\) (B) \(0.49 \mathrm{~cm}\) (C) \(9.8 \mathrm{~cm}\) (D) \(0.98 \mathrm{~cm}\)

Short Answer

Expert verified
The extension produced in the spring is (C) \(9.8 \mathrm{~cm}\).

Step by step solution

01

Convert units

First, it's important to have all units in the same system. Here, we need to convert the natural length of the spring from centimeters to meters. Natural length: \(60 \mathrm{~cm} = 0.6 \mathrm{~m}\)
02

Calculate the force due to gravity on the mass

To find the force acting on the spring due to gravity, we use the formula: Force (\(F\)) = mass (\(m\)) * gravity (\(g\)) Given mass: \(20 \mathrm{~kg}\) and gravity: \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) The force acting on the spring due to mass: \(F = 20 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} = 196 \mathrm{~N}\)
03

Apply Hooke's Law to find the extension

Hooke's Law states that the force acting on a spring is directly proportional to the extension produced in the spring: \(F = k \times x\) Where: - \(F\) is the force acting on the spring (\(196 \mathrm{~N}\)) - \(k\) is the spring constant (\(2000 \mathrm{~N} / \mathrm{m}\)) - \(x\) is the extension produced in the spring (the quantity we need to find) Now, rearrange the equation to solve for the extension \(x\): \(x = \frac{F}{k}\) Plug in the given values of the force and spring constant: \(x = \frac{196 \mathrm{~N}}{2000 \mathrm{~N} / \mathrm{m}}\)
04

Solve for the extension

Perform the division to find the extension: \(x = 0.098 \mathrm{~m}\) Now, convert the extension back to centimeters: Extension: \(0.098 \mathrm{~m} = 9.8 \mathrm{~cm}\)
05

Match the answer to the given options

The calculated extension is \(9.8 \mathrm{~cm}\), which matches option (C). Therefore, the correct answer is: (C) \(9.8 \mathrm{~cm}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: When a gas is allowed to expand, work done by gas is positive. Reason: Force due to gaseous pressure and displacement (of position) on in the same direction. (A) 1 (B) 2 (C) 3 (D) 4

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

A body of mass \(6 \mathrm{~kg}\) is under a force, which causes a displacement in it given by $\mathrm{S}=\left[\left(2 \mathrm{t}^{3}\right) / 3\right](\mathrm{in} \mathrm{m}) .$ Find the work done by the force in first one seconds. (A) \(2 \mathrm{~J}\) (B) \(3.8 \mathrm{~J}\) (C) \(5.2 \mathrm{~J}\) (D) \(24 \mathrm{~J}\)

The coefficient of restitution e for a perfectly elastic collision is (A) 1 (B) 0 (C) \(\infty\) (D) \(-1\)

A body having a mass of \(0.5 \mathrm{~kg}\) slips along the wall of a semispherical smooth surface of radius \(20 \mathrm{~cm}\) shown in figure. What is the velocity of body at the bottom of the surface $?\left(\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(2 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~m} / \mathrm{s}\) (C) \(2 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (D) \(4 \mathrm{~m} / \mathrm{s}\)
See all solutions

Recommended explanations on English Textbooks

English Grammar

Read Explanation

Argumentative Essay

Read Explanation

Pragmatics

Read Explanation

Global English

Read Explanation

Discourse

Read Explanation

Rhetoric

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free