The cable of a 1900-kg elevator has broken, and the elevator is moving downward

Question

The cable of a 1900-kg elevator has broken, and the elevator is moving downward at a steady speed of 1.68 m/s. A safety braking system that works on friction prevents the downward speed from increasing. At what rate is the braking system converting mechanical energy to thermal energy?
while the elevator is moving downward at 1.68 m/s, the braking system fails and the elevator is in free-fall for a distance of 5.02 m before hitting the top of a large safety spring with force constant of 1.50E+4 N/m. After the elevator hits the top of the spring, find the distance d that the spring is compressed before the elevator is brought to rest.

Explanation / Answer

P = F*v = M*g*v = 1900*9.8*1.68 = 3.128x10^4W = 31.3 kW We have (K + U)1 = (K+U)2 where K is kinetic energy and U is potential (m*g*h gravity and 1/2*k*y^2 elastic) Now set Ug = 0 at the contact point of the spring and when it stops K2 = 0 So 1/2*M*v0^2 + M*g*d= 1/2*k*?y^2 So ?y = sqrt((M*v0^2 + 2*M*g*d)/k) = sqrt((1900*1.68^2 + 2*1900*9.8*5.02)/15000) = 3.58 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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