
When a person standing on weighing balance, working on principle of Hooke's law, it shows 60kg after a long time and spring gets compressed by 2.5cm, if a person jumps on balance by 10cm , the maximum reading of balance will be?
A) 60kg
B) 120kg
C) 180kg
D) 240kg
Answer
155.1k+ views
Hint: Hooke's law is a law of physics that states that the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance—that is, F= Kx , where k is a constant factor characteristic of the spring (i.e., its stiffness), and x is small compared to the total possible deformation of the spring . considered spring force and gravitation force to be conservative and gravitational energy is stored into spring by balancing energy net weight to be calculated.
Complete step by step solution:
Let us make following assumptions
Spring constant = K
Mass when just standing =m kg
Mass when jumping = M kg
Extension in spring = x m
According to first condition as per questions when person is just standing
Weight of person is balanced by spring force
Therefore the weight will be 240kg.
Note: Energy stored in spring is conservative hence in ideal conditions taken to be fixed this whole energy is converted into gravitational potential energy when again one of conservative in nature. Energy just converted in another type.
Complete step by step solution:
Let us make following assumptions
Spring constant = K
Mass when just standing =m kg
Mass when jumping = M kg
Extension in spring = x m
According to first condition as per questions when person is just standing
Weight of person is balanced by spring force
Therefore the weight will be 240kg.
Note: Energy stored in spring is conservative hence in ideal conditions taken to be fixed this whole energy is converted into gravitational potential energy when again one of conservative in nature. Energy just converted in another type.
Latest Vedantu courses for you
Grade 11 Science PCM | CBSE | SCHOOL | English
CBSE (2025-26)
School Full course for CBSE students
₹41,848 per year
EMI starts from ₹3,487.34 per month
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Electrical Field of Charged Spherical Shell - JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
