An elevator (mass M 2800 kg) is mov- ing downward at v0 1.35 m/s. A braking system prevents the downward speed from in- creasing.The acceleration of gravity is 9.8 m/s2 .At what rate is the braking system convert- ing mechanical energy to thermal energy?Answer in units of kW.And then while the elevator is moving downward at v0 1.35 m/s, the braking system fails and the elevator is in free-fall for a dis- tance d 2.2 m before hitting the top of a large safety spring with force constant k 26707.3 N/m. After the elevator hits the top of the spring, we want to know the distance ?y that the spring is compressd before the elevator is brought to rest. Writh an algebraic expression for the value of ?y in terms of the known quantities M, v0, g, k, and d, and substitute the given values to find ?y.Answer in units of m.
P F*v M*g*v 2800*9.8*1.35 3.70x10^4W 37.0kW 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((2800*1.35^2 + 2*2800*9.8*2.2)/26707.3) 2.17m
