How do you solve this problem involving springs and compression?

Question

The cable of the 1800 kg elevator shown below snaps when the elevator is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the elevator against guide rails so that a constant frictional force of 4.4 kN opposes the motion of the elevator.
The speed of the elevator before it hits the spring is 7.4 m/s ... The maximum distance the spring is compressed is 0.9 m... The distance the elevator will bounce back up the shaft is 2.756...

Using conservation of energy, find the approximate total distance that the elevator will move before coming to rest in m?

Answer 1

initial energy = final energy



mgS - ½ m V² - f S = 0



(1800)(9.8)S - ½ (1800)(7.4)² - (4400) S = 0



S = 3.72 m



let X displacement of spring due to impact by elevator,



initial energy = final energy



mg(X + S) - ½ k X² - f (X + S) = 0



(1800)(9.8)(X + 3.72) - ½ (0.15 E6) X² - 4400 (X + 3.72) = 0



solve quadratic equation above, and choose positive value for X, we get:



X = 0.9 m...... ..(OK appropriate with question above)



let Y is distance of elevator bouncing,



initial energy = final energy



½ k X² - f Y - m g Y = 0



½ (0.15 E6)(0.9)² - (4400) Y - 1800(9.8)Y= 0



Y = 2.756 m...... ..(OK appropriate with question above)



ratio = Y/(X + S) = 2.756/(0.9 + 3.72)



ratio = 0.596



total distance during elevator moving until rest is



Z = (X + S) + 2Y/(1 - ratio) = (0.9 + 3.72) + 2(2.756)/(1 - 0.596)



Z = 18.27 m

Answer 2

Initial potential energy = Final potential energy + Work performed



Because the spring is unloaded at t=0, the only potential energy in the system is gravitational potential energy in the elevator. Using the unloaded location of the spring as the origin,



Initial potential energy = mgd



When the elevator comes to rest, the potential energy is stored in the spring and in the elevator. Using conservation of external forces on the elevator at rest on the spring, you find the final resting height of the elevator. Choosing downward force as negative to keep the signs straight,



-mg - kX = 0, where X = final resting height



Final potential energy = mgX + ½kX²



Work performed is the work overcoming friction over the total travel of the elevator. Because the frictional force is constant and directly opposes the elevator,



Work performed = FD, where F is the frictional force due to the safety device and D is the total travel of the elevator



mgd = mgX + ½kX² + FD



-mg - kX = 0 therefore X = -mg/k (negative indicating compressed spring)



mgd = mg * (-mg/k) + ½k(-mg/k)² + FD



mgd = -m²g² / k + ½m²g² / k +FD



D = (mgd + ½m²g²/ k) / F



Substituting known values,



D = (1800 * 9.8 * 3.7 + ½ * 1800² * 9.8² / 150000) / 4400



D = 15.07 m

Answer 3

I assuming you recognize a thank you to transform instruments. it incredibly is an potential conservation subject. you have 3 issues occurring the spring, gravity, and the bullet shown so as below. a million/2*ok*d^2 + M*g*d = a million/2*m*Vb^2 remedy for Vb in this... Vb = [(ok*d^2 + 2*M*g*d) / m]^0.5 then you definately in simple terms plug on your numbers and verify the instruments. Vb^2 = [(50 kg*m/m*s^2)*(.45m)^2 + 2*(2 kg)*(9.80 one m/s^2)*(.45m)] / 0.a million kg = 277.80 3 m^2 / s^2 Take the sq. root and you get Vb = [277.86 m^2 / s^2]^0.5 = sixteen.668 m/s