JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

Two Block Spring System

Here, we will understand the mechanism of the two block-spring system. We will look at an experiment and understand all the related terms, as well as learn to solve problems.

Two-block Spring System Experiment and Mechanism

A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane. At first, the blocks are at rest, and the spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Now, we have to find the maximum extension of the spring.

Two Block Spring System Mechanism

Here, we take the two blocks plus the spring as a system.

Since there is an external force F on the system, there will be some acceleration of the centre of mass, which can be calculated by the formula

\(\begin{array}{l}\overrightarrow{a_{cm}} = \frac {\overrightarrow{F_{ext}}}{M}  \end{array} \)

Where,

\(\begin{array}{l}\overrightarrow{F_{ext}}\ \text{is the net external force on the system}\end{array} \)

M is the total mass of the system.

\(\begin{array}{l}a = \frac {F}{m ~+~ M} \end{array} \)

Now, let us solve this problem from the frame of reference of a centre of mass. This means that we would have to apply a pseudo force ma towards the left of the block of mass m and another pseudo force Ma towards the left on the block of mass M.

Net external force on the block of mass m is

\(\begin{array}{l}F_1 = ma = \frac {mF}{m + M}\end{array} \)

This force will be in the left direction.

Net external force on the block of mass M is

F2 = F – Ma

\(\begin{array}{l}=F – \frac {MF}{m + M} = \frac {mF}{m~+ M} \end{array} \)

This force will be in the right direction.

The situation from the frame of the centre of mass would be as shown in the picture below.

Two Block Spring System

The centre of mass is at rest in this frame, the blocks will move in opposite directions, and after some time, when the extension of the spring is maximum, they will come to rest for an instant. Let the block of mass m moves a distance x1, and the block of mass M moves a distance x2. The total work done by forces F1 and F2 will be

W = F1 x1 + F2 x2

\(\begin{array}{l}=\frac {mF}{m + M} ( x_1 + x_2) \end{array} \)

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

\(\begin{array}{l} \frac{mF}{m+M} (x_1+x_2) = \frac 12 k ( x_1+x_2)^2\end{array} \)

Or

\(\begin{array}{l}( x_1+x_2) = \frac {2 mF}{k ( m+M)}\end{array} \)

Recommended Video

Understanding Spring Constant – Applying Concepts

If you want to learn more about spring block systems, connect with our mentors, here at BYJU’S classes.

Frequently Asked Questions on Spring Block System

Q1

Define the force constant.

The force constant is defined as the restoring force per unit displacement.

Q2

What provides the restoring force when a spring is compressed and then left free to vibrate?

The elasticity of the material of the spring.

Q3

What unit does the spring constant k have?

N/m or kg/m2

Test Your Knowledge On Spring Block System!

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