# The system shown in the figure is in equilibrium at rest and the spring and string are massless. Now the string is cut. The acceleration of masses 2m and m just after the string is cut will be:

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Updated on: September 25th, 2023

The acceleration of masses 2m and m just after the string is cut will be g/2 upwards, g downwards

When the system is in equilibrium:

FBD of the system:

applying the vertical equilibrium condition to the system

Fspring = 3mg

After the string is cut, T = 0 i.e tension in string becomes zero, while spring force acts at its initial value Fspring, because of inertia of spring.

FBD of the individual blocks following string cutting is shown here:

Acceleration of block of mass m is given by

am = mg/m

On simplifying we get

am = g (Downwards)

Acceleration of block of mass 2m is given by

a2m = Fspring – 2mg/ 2m = mg/2m

On simplifying we get

a2m = g/2 (upwards)

### Features of Free Body Diagram

A diagram that changes when the issue is resolved is called a free-body diagram. A free body diagram often includes the following elements:

• a condensed form of the body (most commonly a box)
• a coordinate framework
• Arrows showing in the direction in which forces act on the body are used to depict forces.
• Moments appeared as curved arrows pointing towards the body’s acting direction.

The specific problem and the presumptions used will determine how many forces are acting on a body. Friction and air resistance are frequently ignored.

Summary:-

## The system shown in the figure is in equilibrium at rest and the spring and string are massless. Now the string is cut. The acceleration of masses 2m and m just after the string is cut will be:

The system shown in the figure is in equilibrium at rest and the spring and string are massless. Now the string is cut. The acceleration of masses 2m and m just after the string is cut will be g/2 upwards, g downwards

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