A 1.02 kg mass is attached to a spring of force constant 10.2 N/cm and placed on

Question

A 1.02 kg mass is attached to a spring of force constant 10.2 N/cm and placed on a frictionless surface. By how much will the spring stretch if the mass moves along a circular path of radius 0.480 m at a rate of 2.02 revolutions per second?

Explanation / Answer

omega = 2 x pi x n = 2 x 3.14 x 2.14 = 13.44 rad/sec =>v = r x omega =>v = 0.485 x 13.44 = 6.52 m/s F = mv^2/r = kx =>x = mv^2/rk = (1.01 x 6.52 x 6.52)/(0.485 x 9.5) = 9.32m s = stretch of the spring F = force applied k = spring constant = 894.723 Newtons/m F = ks (b) F = 1.4 x 9.81 = 13.734 newtons s = F/k = 13.734/894.723 = 0.01535 m = 1.535 cm (c). Work = force x distance F =( 9.2/100) x 894.723 = 82.3145 Newtons W = 82.3145 x 9.2/100 = 7.573 Newton - meters

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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