Question

Light helical spring obeying Hooke’s Law was attached to a fixed support. When a 5N load was hung on it, the length of the spring was 160mm. when a 10N load was hung on it the length became 200mm. determine

1. The length of the spring with no load.
2. The length of the spring with 8N.

Answer 1

1. F=ke
   5= K(0.160 – l_o)
   10 = K(0.2 – l_o)
   
   The formula may be implied.
   5/10=0.16l_o/0.02l_o
   1.5 l_o = 1.6 - 10 l_o
   (10 – 5) l_o = 1.6 – 1
   5 l_o = 0.6
   l_o = 0.12m
   = 12cm
    
2. F= Ke
   10 = K X 0.08
   K = ¹⁰/_0.08=¹⁰⁰⁰/₈=125N/m        
   E= f/k    =8/125 x 100 =6.4cm
   L = L_o + e
   = 12+ 6.4 = 18.4cm