**Question-**

A block mass **m = 1kg** is placed over another block of mass **M = 3kg**. **M** is placed over a **smooth surface**. Both the blocks move with **velocity v** towards a spring of constant **K = 400 N/m** and

**μ = 0.2** for the two blocks. Then, the maximum value of **v**, so that there may not be slipping between **m** and **M** is -

**My Attempt-**

Let me call block **m** and block **M** together as the **system**. When the system compresses the spring and momentarily comes to rest, the **compression** can be found out through conservation of energy.

**Elastic Potential Energy Gained = Kinetic Energy Lost (Work Done by Friction is Internal)**

**1/2 Kx ^{2} = 1/2 (M+m) v^{2}**

Plugging in the values,

**x = v/10**

During this compression,

**Kinetic Friction Force**acts on block m and reduces its velocity to

**zero**.

**Deceleration**due to Kinetic Friction Force is given by,

**2N = 1kg x a**

**a = 2m/s**

^{2}Since this is the

**maximum frictional force**that is acting, the

**maximum velocity**that it can reduce to zero over the

**same distance x**can be obtained from

**kinematics**.

On

**substituting x = v/10**in the kinematics equation, we get

**v = 0.4 m/s**

I am told that this answer is wrong. **Have I committed any mistake?**