A block of mass m has initial velocity u having direction +x axis. The block stops after covering distance S causing similar extension in the spring of spring constant K holding it. If μ μ is the kinetic friction between the block and the surface on which it was moving, the distances S is given by
A
μ μ 2 m 2 g 2
B
(mKu 2 - μ μ 2 m 2 g 2 ) 1/2
C
( μ μ 2 m 2 g 2 + mKu 2 + μ μ mg) 1/2
D
( μ μ 2 m 2 g 2 – mKu 2 + μ μ mg) 1/2
Text Solution
Verified by ExpertsThe correct answer is:
D
mu 2 = - μ μ mgS + 
KS 2
i.e., S 2 + 
Solving for μ μ ,
S = 
( μ μ 2 m 2 g 2 – mKu 2 - μ μ mg)
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