A block of mass 2 kg is tied to spring of natural length 2 m whose other end is fixed at point O. Initially, the block is at same level as that of O with spring unstretched. Now block is released and its speed is 5 m/s when the spring is in vertical position. If the acceleration of the block at this lowest position is (5/2)K m/s², find the value of K.
Physics · Mixed Chapter
JEE Mains

Question

A block of mass 2 kg is tied to spring of natural length 2 m whose other end is fixed at point O. Initially, the block is at same level as that of O with spring unstretched. Now block is released and its speed is 5 m/s when the spring is in vertical position. If the acceleration of the block at this lowest position is (5/2)K m/s², find the value of K.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: From the diagram, the vertical spring length at the lowest point is 4 m, so extension is x=4−2=2 m and the block drops h=4 m. Step 2: Energy conservation: mgh=½mv²+½kx². With m=2 kg, g=10 m/s², v=5 m/s: 80=25+2k, giving k=27.5 N/m. Step 3: At the lowest point, vertical spring force is kx=55 N up, gravity is mg=20 N down. Net acceleration magnitude is (55−20)/2=17.5 m/s². Step 4: 17.5=(5/2)K, hence K=7.