A spring with a force constant of 3.0×10 4 N/m is initially at its equilibrium l

Question

A spring with a force constant of 3.0×104 N/m is initially at its equilibrium length. How much work must you do to stretch thespring 0.070 m? How much work must you do to compress it0.070 m? I thought you would be using theequation: W=1/2KX^2 but when i plug it in,, say forcompression. W=1/2(3000N/m)(.070m)^2=7.35 it tells me itswrong. What am i doing wrong? How do i solve for the stretch? A spring with a force constant of 3.0×104 N/m is initially at its equilibrium length. How much work must you do to stretch thespring 0.070 m? How much work must you do to compress it0.070 m? I thought you would be using theequation: W=1/2KX^2 but when i plug it in,, say forcompression. W=1/2(3000N/m)(.070m)^2=7.35 it tells me itswrong. What am i doing wrong? How do i solve for the stretch? How much work must you do to stretch thespring 0.070 m? How much work must you do to compress it0.070 m? I thought you would be using theequation: W=1/2KX^2 but when i plug it in,, say forcompression. W=1/2(3000N/m)(.070m)^2=7.35 it tells me itswrong. What am i doing wrong? How do i solve for the stretch?

Explanation / Answer

given force constant is K = 30000 N/m so the work done is = 1/2 (30000)(0.07)2 = 73.5J for the strech also we get the same work you left one zero in force constant

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.