Question

A pen contains a spring with a spring constant of 275 N/m. When the tip of the pen is in its retracted position, the spring is compressed 5.6 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.7 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer. ***Please show detail work***

Answer 1

initial compression in spring dx1 = 5.5 mm = 5.6*10^-3 m

distance moved from mean position x1 = 5.5 mm = 5.6*10^-3 m

energy stored U1 = (1/2)*k*x1^2

to ready the pen, the spring is compressed by additiona distance dx2 = 6.7 mm

distance moved from mean position x2 = 5.5 + 6.7 = 12.2 mm = 12.2*10^-3 m

energy stored U2 = (1/2)*k*x2^2

work done = U2 - U1 = (1/2)*k*(x2^2 - 1^2)

W = (1/2)*275*((12.2*10^-3)^2 - (5.6*10^-3)^2)

W = 0.0162 J <<<------ANSWER