# A vertical spring stretches 10 cm under a load of 200 g. What is the spring constant, how much work is required to stretch the first 5 cm and the last 5 cm.

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### 1 Answer

The vertical spring has a load 200 g suspended from it. It stretches 10 cm under the load. If the spring constant is k, the potential energy stored in the spring when it is extended by x is given as (1/2)*k*x^2. The decrease in gravitational potential energy is equal to m*g*h = 0.2*9.8*0.1 = 0.196. This energy is converted to the potential energy stored in the spring. The two can be equated to give:

(1/2)*k*0.1^2 = 0.196

=> k = 39.2 N/m

The work required to stretch the first 5 cm is 0.5*39.2*0.5^2 = 0.049 J. The work required for the next 5 cm is 0.196 - 0.049 = 0.147 J

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