A 1.1-kg object is suspended from a vertical spring whose spring constant is 150 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is then pulled straight down by an additional distance of 0.18 m and released from rest. Find the speed with which the object passes through its original position on the way up.

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

where;

v is the final speed of the object at its original position;

Therefore, the speed with which the object passes through its original position on the way up is 2.94 m/s.

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explanation:

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