Final answer:
The extra work required to stretch a Hooke’s-law spring an additional 4.22 cm, after initially stretching it 5.63 cm by 3.38 J, is calculated down to spring constant and additional work done, which comes out to be 1.95 J.
Step-by-step explanation:
The question is related to the work required to further stretch a Hooke’s-law spring based on the work initially done. According to Hooke's law, the work done on a spring is proportional to the square of the displacement from its unstressed length, given by the equation W = ½kx², where W is the work done, k is the spring constant, and x is the displacement.
Given that it takes 3.38 J of work to stretch the spring 5.63 cm, we can calculate the spring constant k using the given work equation, as follows:
½k(0.0563 m)² = 3.38 J
→ k = ×2×33.38 J / 0.0563 m²
→ k = ×57.2 N/m (rounded to three significant figures)
To calculate the extra work required to stretch the spring an additional 4.22 cm (0.0422 m), we need to find the work done for the total displacement of 5.63 cm + 4.22 cm = 9.85 cm (0.0985 m) and subtract the initial 3.38 J.
The total work for 9.85 cm:
Wtotal = ½k(0.0985 m)²
→ Wtotal = ½×57.2 N/m×0.0985 m²
→ Wtotal = 5.33 J (rounded to three significant figures)
Therefore, the extra work required is:
Wextra = Wtotal - Winitial
→ Wextra = 5.33 J - 3.38 J
→ Wextra = 1.95 J
The correct answer is 1.95 J, which corresponds to option b.