The cable of the 2200 kg elevator cab in the figure snaps when the cab is at res

Question

The cable of the 2200 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 3.1 m above a spring of spring constant k = 0.15 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 3.9 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft.(d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

Explanation / Answer

Help from this

The cable of the 2200 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 4.4 m above a spring of spring constant k = 0.36 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 3.2 kN opposes the cab's motion

Ans

This is not hard at all but going on your going to use this equation

the change in kinetic energy + the change in potential energy = -( the friction times the distance) + all other external forces

we have no other external forces, no one is pushing this thing everything is internal towards the system

so

k stands for kinetic energy and u stands for potential
i and f for initial and final

kf - ki + uf - ui = -fkd

no initial kinetic energy the object is at rest, and no final potential all the energy has been converted to kinetic energy

so

kf = -fkd + ui

0.5mv^2 = -fkd + mgh

0.5 (2200) v^2 = 3.2kN(1000N/kN)4.4m + 2200 * 9,8 * 4.4

solve for v

part b

the maximum distance comprised equals

kf - ki + uf - ui = -fkd

our final kinetic energy is zero and our initial spring & gravitational energy is zero,

uspringfinal = -fkd + ki - ugravpotential

0.5 k x^2 = -fkx + 0.5*mass*velocity(use the velocity from part a) - mg(-x)

if we define our origin at the maximum height of the spring we will have a negative x for our distance for gravity, we used this for our advantage to get the answer part a

0.5(0.36MN/m)(convert into newtons) x ^ 2 -mg(x) +fkx - 0.5 * mass * velocity = 0

0.5(0.36)(convert into newtons, i forgot what to multiply) x^2 -(2000*9.8 + 3.2 * 1000) x - 0.5*(2000)(v)^2 = 0

your going to have use the quadratic equation to solve for x , pick the positive x

part c

using the x you found from part b

once again assuming friction is still acting (if not just omit fkd)

kf - ki + uf - ui = -fkd

kf & ki are both zero, zero when at maximum compression and zero at maximum height

we have an initial gravitational potential energy and spring potential energy, no final spring potential energy (the elevator is above the spring , the spring is no longer at it, the point of maximum is our potential final height)

mgh = -fkd + uspring initiail - mg(-x)

plug in the x you solved for

mgh + fkd = 0.5kx^2 + mg(x)

h = d the friction force is constant throughout the travel

h (mg + fk) = 0.5kx^2 + mg(x)

solve for h all other values should be known by now

d)

know you going to solve for the d in the friction force

you have final kinetic and spring potential, and initial gravitational

0.5mv^2 + 0.5kx^2 - mg(initial height) = -3.2*1000 * distance

solve for distance

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