Answer:
The speed with which the object passes through its original position on the way up is 2.94 m/s.
Explanation:
Given;
mass of the object suspended, m = 1.1 kg
spring constant, k = 150 N/m
(a)the amount by which the spring is stretched from its unstrained length;
F = Kx
mg = Kx
x = mg / K
where;
x is the extension of the spring
x = (1.1 x 9.8) / 150
x = 0.072 m
x = 7.2 cm
(b) when the object is pulled straight down by an additional distance of 0.18 m;
total extension = 0.072 m + 0.18 m = 0.252 m
Elastic potential energy of the spring at this position is given by;
U = ¹/₂Kx²
U = ¹/₂ x 150 x (0.252)²
U = 4.7628 J
Based on law of conservation of mechanical energy, this potential energy of the spring will be converted to kinetic energy of the object;
K.E = U
¹/₂m(v-u)² = U
¹/₂m(v-0)² = U
¹/₂mv² = U
v² = 2U /m
[tex]v = \sqrt{\frac{2U}{m} }[/tex]
where;
v is the final speed of the object at its original position;
[tex]v = \sqrt{\frac{2*4.7628}{1.1} }\\\\v = 2.94 \ m/s[/tex]
Therefore, the speed with which the object passes through its original position on the way up is 2.94 m/s.
