question_answer

Read the following passage and answer the question given below: One end of a spring is attached to a fixed vertical support and the other end to a block which can move on a horizontal table. Let \[x=0\] denote the position of the block when the spring is in its natural length. When the block is displaced by an amount\[x\] (either compressed or elongated) a restoring force(F) is applied by the spring on the block. The direction of this force F is always towards its mean position \[(x=0)\] and the magnitude is directly proportional to\[x\]or \[F\propto x\]or\[F=-kx\]where k is a constant called force constant of spring and depends on the nature of the spring. A particle of mass 0.3 kg is subjected to a force \[F=-kx\] with k = 15 N/m. What will be its initial acceleration if it is released from a point\[x=\text{2}0\text{cm}\]?

Answer:

Given m = 0.3 kg, Spring constant \[K=15N/m\] elongation \[x=20cm=\frac{20}{100}m,\,\,a=?\] We know, \[F=-kx\] \[\Rightarrow ma=-ka\,\,\Rightarrow 0.3\times a=-15\times \frac{1}{5}\] \[=a=\frac{3}{0.3}=10m/{{s}^{2}}\]