Answer:The spring force is conservative, so the work done by the spring force is equal to the negative of the potential energy stored in the spring:
U = -1/2 k x^2
where k is the spring constant and x is the displacement from the unstrained length.
The initial compression of the spring is 4.10 mm = 0.00410 m, and the additional compression is 6.10 mm = 0.00610 m. The total compression of the spring is therefore x = 0.00410 m + 0.00610 m = 0.0102 m.
The potential energy stored in the spring when it is compressed by a distance x is:
U = -1/2 k x^2
Substituting the given values, we get:
U = -1/2 (216 N/m) (0.0102 m)^2
U = -0.0112 J
The work done by the spring force to ready the pen for writing is equal to the change in potential energy:
W = U_final - U_initial
where U_initial is the potential energy of the spring when it is compressed 4.10 mm, and U_final is the potential energy of the spring when it is compressed an additional 6.10 mm.
U_initial = -1/2 (216 N/m) (0.00410 m)^2 = -0.000090 J
U_final = -1/2 (216 N/m) (0.0102 m)^2 = -0.0112 J
W = U_final - U_initial
W = (-0.0112 J) - (-0.000090 J)
W = -0.0111 J
The negative sign indicates that the work done by the spring force is done on the pen (i.e. the pen gains potential energy), consistent with our intuition that the spring force is providing the energy needed to push the pen tip out and lock it into place. Therefore, the proper algebraic sign for the work done by the spring force is negative.
Step-by-step explanation: