A force of 10N extends a spring of natural length 1m by 0.02m, calculate the length of the spring when the applied force is 40N.

Assessment: JAMB UTME - Physics - 2024 Subject: Physics

Question 1 Report

A force of 10N extends a spring of natural length 1m by 0.02m, calculate the length of the spring when the applied force is 40N.
Answer Details

To solve this problem, we will use Hooke's Law. Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is represented as:


F = k * x


where:

  • F is the force applied to the spring.
  • k is the spring constant.
  • x is the extension or compression of the spring from its natural length.

Firstly, we need to find the spring constant k. We know that a force of 10N extends the spring by 0.02m. Therefore, using Hooke's Law:


10N = k * 0.02m


From this, we can solve for k:


k = 10N / 0.02m = 500N/m


Now that we have determined the spring constant, let's calculate the extension caused by a force of 40N:


Using Hooke's Law again:


F = k * x

40N = 500N/m * x


Solving for x:


x = 40N / 500N/m = 0.08m


This means that the spring is extended by 0.08m when a force of 40N is applied. Therefore, the length of the spring (natural length plus extension) becomes:


1.00m + 0.08m = 1.08m


Thus, the **length** of the spring when the applied force is 40N is 1.08m.

Download The App On Google Playstore

Everything you need to excel in your exams

Green Bridge CBT Mobile App
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning