A force of magnitude 800N stretches a certain spring by 0.200mform its equilibri

Question

A force of magnitude 800N stretches a certain spring by 0.200mform its equilibrium position. a)    what is the force constant of thisspring?          k= N/m b) how much elastic potential energy is stored in the springwhen it is stretched 0.300m from its equilibriumposition.?      U = J c) how much elastic potential energy is stored in the springwhen it is compressed by 0.300m from its equilibrium?    U = J d) how much work was done is streching the spring by theoriginal 0.200m?    W = J A force of magnitude 800N stretches a certain spring by 0.200mform its equilibrium position. a)    what is the force constant of thisspring?          k= N/m b) how much elastic potential energy is stored in the springwhen it is stretched 0.300m from its equilibriumposition.?      U = J c) how much elastic potential energy is stored in the springwhen it is compressed by 0.300m from its equilibrium?    U = J d) how much work was done is streching the spring by theoriginal 0.200m?    W = J

Explanation / Answer

(a): F = kx 800 = k(0.200) k = 800/0.200 k = 4000 N/m -- (b) U = 1/2kx^2 U = (0.5)(4000)(0.300)^2 U = 180 J -- (c) same distance in different direction, so same magnitde indifferent direction = 180 J -- (d) Work = F*d Work = (0+800)/2*0.200 Work = 80 J -- Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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