Understanding a Spring Block System for the IIT JEE Entrance Exam

Qualifying the IIT JEE entrance exam is of course not a bed of roses. The exam comprises of two stages – JEE Main and JEE Advanced. The top 2,50,000 candidates who qualify the JEE Main exam are eligible to appear for the JEE Advanced exam which is conducted for admission to prestigious IITs. Thus, students need to work very hard to get a good rank in these entrance examinations. Both the exams comprise of questions from Physics, Chemistry, and Mathematics. Physics is one of the subjects in which many theoretical and numerical problems are asked. The spring block system is one of the most important concepts that students need to prepare by putting their hearts and souls into it. Here, we are providing a brief about the topic.

Latest Update for JEE Entrance Exams The JEE Main exam conducting authority (NTA) has postponed the April session of the exam due to COVID-19 situation. Subsequently, IIT Delhi has postponed the JEE Advanced exam. The JEE Advanced exam is conducted only after the JEE Main result. The new dates for both the exams will be announced soon.

What Is Spring Block System?

Let a block of mass ‘m’ be connected to another block of mass ‘M’ through a massless spring of spring constant ‘k’. Keep the blocks on a smooth horizontal plane. Let them be at rest first. Now, stretch the spring with a constant force ‘F’ on the block of mass ‘M’ in order to pull it. Now, calculate the maximum extension of the spring.

Considering the two blocks and the spring as a system

As there is an external force ‘F’ acting on the system, there will be an acceleration in the centre of mass, which can easily be calculated by the formula:

Overrightarrowacm=fracoverrightarrowFextM

where, OverrightarrowFext is the net external force on the system

M is the total mass of the system

a = fracFm + M

Now, see solve this complicated problem from the frame of reference of a centre of mass. This means that you will have to apply a pseudo force ‘ma’ towards the left of the block of mass ‘m’ and another pseudo force ‘Ma’ towards left on the block of mass ‘M’.

Total external force on the block of mass ‘m’ is given by:

F1 = ma = fracmFm + M

Note: This force will be in the left direction

Total external force on the block on mass ‘M’ is given by:

F2 = F – Ma

= F – fracMFm + M=fracmFm + M

This force will be in the right direction

Now, let the block of mass ‘m’ move a distance X1 and the block of mass ‘M’ move a distance X2.

The total work done by these forces F1 and F2 will be as follows:

W = F1 x1 + F2 x2

= fracmFm+M(x1 + x2)

This amount of work done should be equal to the change in potential energy of the spring as the change in kinetic energy will be zero from the frame of centre of mass.

fracmFm + M(x1 + x2) =frac12k(x1 + x2)2

Or

(x1 + x2)=frac2mFk (m + M)

Students should note that numericals related to a spring block system are very tough. They should try to understand it properly. They should lay focus on the crux of the problems.

Questions asked in JEE:

Questions related to a spring block system will be of objective type. Students are required to tick the right answer in the shortest possible time. They should not try their best guesses as there is negative marking in the JEE exam. With best practices, they should solve the problems and get a solution. Practice the questions at the back of the topic and in sample papers and previous years’ question papers. The more you will practice the questions, the better you will get the topic. Consistent practice is the only key to solve complex questions in the JEE exam’s Physics section.

Updated: May 4, 2020 — 9:22 am

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