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A block of mass m is attached to a massless spring of spring constant k. This system is accelerated upward with acceleration a. The elongation in spring will be?
Most Upvoted Answer
A block of mass m is attached to a massless spring of spring constant ...
Analysis:
The motion of the block attached to the spring can be analyzed using Newton's second law and the spring force.

Newton's Second Law:
The net force acting on the block is given by:
\[ F_{\text{net}} = ma \]

Spring Force:
The spring force is given by Hooke's Law:
\[ F_{\text{spring}} = -kx \]
where x is the elongation in the spring.

Equilibrium of Forces:
At equilibrium, the net force is zero:
\[ ma = kx \]
Solving for x, we get:
\[ x = \frac{ma}{k} \]
Therefore, the elongation in the spring when the system is accelerated upward with acceleration a is \[ \frac{ma}{k} \].
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A block of mass m is attached to a massless spring of spring constant ...
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A block of mass m is attached to a massless spring of spring constant k. This system is accelerated upward with acceleration a. The elongation in spring will be?
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