A 2.30 block on a horizontal floor is attached to a horizontal spring that is in

Question

A 2.30 block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0340 . The spring has force constant 890 . The coefficient of kinetic friction between the floor and the block is 0.36 . The block and spring are released from rest and the block slides along the floor
Part A
What is the speed of the block when it has moved a distance of 0.0160 from its initial position? (At this point the spring is compressed 0.0180 .)
Express your answer with the appropriate units.

Explanation / Answer

intial energy of the system = energy store in spring = (1/2)*k*x2 = (1/2)*890*(0.034)2 = 0.51442 J

final energy = KE + energy in spring

                     = KE + (1/2)*890*(0.018)2 = KE+0.14418

work done against friction = frictional force*distance = *m*g*s = 0.36*2.3*9.8*(0.016) = 0.1298 J

intial energy of system- work against friction = final energy

0.51442 - 0.1298 = KE + 0.14418

KE = 0.24044 J

(1/2)*2.3 * (v2 ) = 0.24044

v =0.45725 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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