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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:
By BYJU'S Exam Prep
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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