Blocks m₁=2 kg and m₂=8 kg hang over an ideal pulley. Springs k₁=50 N/m and k₂=70 N/m connect the respective masses to the ground as shown. The second spring is compressed 0.5 m. Find x if first spring's deformation is x/10 m.
Physics · Mixed Chapter
JEE Main Practice

Question

Blocks m₁=2 kg and m₂=8 kg hang over an ideal pulley. Springs k₁=50 N/m and k₂=70 N/m connect the respective masses to the ground as shown. The second spring is compressed 0.5 m. Find x if first spring's deformation is x/10 m.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: On the 8-kg mass, assuming g=10 m/s² and spring 2 compressed upward, equilibrium gives T +70(0.5)=80 ⇒T=45 N. Step 2: On the 2-kg mass, equilibrium requires T+F_s1 = 20 ⇒F_s1 =−25 N, indicating spring 1 acts downward (extension) rather than upward compression. Step 3: Magnitude of deformation is |F_s1|/k₁ =25/50=0.5 m ⇒ x=5. Step 4: Although magnitude matches source key, the first spring should be described as extended, not compressed, for the drawn geometry.