The 2.6-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of kinetic friction is 0.53. Calculate (a) the velocity v of the collar as it strikes the spring and (b) the maximum deflection x of the spring.

The attached diagram explains the system,

Sum of Fy = 0

N=9.81

N - mgCos60 = 0

F= ukN= (0.53)(9.81) =

F= 5.12 N

So

F.d= 1/2(mv.v) - mgdsin60

-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)

(a)  v = 2.436 m/s

For deflection

-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)

by solving for with values of v, m, g, F, k

800x^2 - 11.87 x - 5.938 = 0

by solving the quadratic equation

x = 0.093, -0.079

(b) x = 0.093 m

correct Answer is 0.093m

Explanation:

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