A spring of spring constant k = 350 Nm -1 is used tolaunch a 2.5-kg block along

Question

A spring of spring constant k = 350 Nm-1 is used tolaunch a 2.5-kg block along a horizontal surface by compressing thespring by a distance of 22 cm. If the coefficient of slidingfriction between the block and the surface is 0.50, how far doesthe block slide?

Explanation / Answer

We know that workdone by frictional force = change inpotential energy       ==>             fk * s = 0.5kx2       ==>            k mg * s = 0.5kx2       ==>                          s = 0.5 kx2 / k mg                                           = 0.5 * 350 * 0.22*0.22 / 0.5 * 2.5 * 9.8                                           = 0.6914 m                                           = 69.14 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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