Question

A pen contains a spring with a spring constant of 292 N/m. When the tip of the pen is in its retracted position, the spring is compressed 4.40 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 6.00 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.

Answer 1

Spring constant K = 292 N/m

At retracted position ,compression X = 4.4 mm = 4.4 x10 -3 m

Potentila energy of the spring U = (1/2) KX ²

= 0.5 x 292x(4.4x10 ^-3) ²

= 2.82656x10 ^-3 J

the spring must be compressed an additional 6.00 mm:

Total Compression X ' = 4.4 mm + 6 mm= 10.4 mm = 10.4 x10 -3 m

Potentila energy of the spring U ' = (1/2) KX ' ²

= 0.5 x 292x(10.4x10 ^-3) ²

= 15.79x10 ^-3 J

work is done by the spring force to ready the pen for writing W = U - U '

= -12.9648x10 ^-3 J