One end of a light spring of natural length d and spring constant k = 64 N/m is fixed on a rigid wall and the other is attached to a smooth ring of mass m = 1 kg which can slide without friction on a vertical rod fixed at a distance d = 3 m from the wall. Initially the spring makes an angle 37° with the horizontal as shown in fig. When the system is released from rest find the speed of the ring when the spring becomes horizontal (in m/s) [sin 37° = 3/5].
Physics · Mixed Chapter
JEE Mains

Question

One end of a light spring of natural length d and spring constant k = 64 N/m is fixed on a rigid wall and the other is attached to a smooth ring of mass m = 1 kg which can slide without friction on a vertical rod fixed at a distance d = 3 m from the wall. Initially the spring makes an angle 37° with the horizontal as shown in fig. When the system is released from rest find the speed of the ring when the spring becomes horizontal (in m/s) [sin 37° = 3/5].

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: cos37°=4/5, so the initial spring length is d/cos37°=3.75 m and extension is 3.75−3=0.75 m. Step 2: Initial vertical height above the horizontal-spring position is h=d tan37°=3×3/4=2.25 m. Step 3: At the horizontal position the spring reaches natural length. Energy conservation gives (1/2)mv²=mgh+(1/2)k(0.75)². Step 4: (1/2)v²=22.5+18=40.5, so v²=81 and v=9 m/s.