In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00 {\rm m/s} when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 {\rm m} as it does so. During the motion a safety clamp applies a constant 17000-{\rm N} frictional force to the elevator.
What is the speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?
conservation of energy.
KE + PE --> Wf + PEspring
.5mv^2 + mgh = 17000*1 + .5kx^2
v = 3.65
To find the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts the spring, we need to calculate the work done on the elevator by the various forces (gravity, spring, and friction) in this distance.
First, let's find the gravitational potential energy of the elevator at the initial point of contact with the spring:
Gravitational potential energy = mgh
where m is the mass of the elevator (2000 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height (distance fallen) before contacting the spring. Since the elevator is falling, h is given by the speed (v) and acceleration due to gravity:
h = (1/2)gt^2
where t is the time taken to reach the point of contact with the spring.
To find t, we can use the equation of motion:
v = u + at
where u is the initial velocity (4.00 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and v is the final velocity at the point of contact (which is 0 m/s since the elevator starts to compress the spring).
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (0 - 4.00) / (-9.8)
t = 4.00 / 9.8
t ≈ 0.41 s
Using this value of t, we can calculate h:
h = (1/2)(9.8)(0.41)^2
h ≈ 0.083 m
Now that we have the initial height, we can calculate the initial gravitational potential energy:
Gravitational potential energy = mgh
Gravitational potential energy = (2000)(9.8)(0.083)
Gravitational potential energy ≈ 1625 J
Next, let's consider the work done by the cushioning spring. The spring compresses by 2.00 m, so the work done by the spring can be calculated using Hooke's Law:
Work done by the spring = (1/2)kx^2
where k is the spring constant and x is the compression distance.
The spring constant is not given directly, but we can find it using the given information. Since the spring is supposed to stop the elevator, the spring force (kx) is equal to the weight (mg) of the elevator:
kx = mg
k = mg / x
Substituting the given values:
k = (2000)(9.8) / 2.00
k ≈ 9800 N/m
Now that we have the spring constant, we can calculate the work done by the spring:
Work done by the spring = (1/2)(9800)(2.00)^2
Work done by the spring = 19,600 J
Finally, let's consider the work done by the frictional force. The frictional force is constant (17,000 N), and the displacement is 1.00 m. The work done by a constant force is given by the product of the force and the displacement:
Work done by friction = Frictional force × displacement
Work done by friction = 17,000 × 1.00
Work done by friction = 17,000 J
To find the final kinetic energy of the elevator, we subtract the work done by the friction and the work done by the spring from the initial gravitational potential energy:
Final kinetic energy = Initial gravitational potential energy - Work done by friction - Work done by the spring
Final kinetic energy = 1625 J - 17,000 J - 19,600 J
Final kinetic energy ≈ -35,975 J
Since the kinetic energy cannot be negative, we assume it is zero, as the elevator comes to a stop after compressing the spring. Therefore, the speed of the elevator after moving downward 1.00 m from the point of first contact with the spring is 0 m/s.
To find the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring, we can use the principle of conservation of mechanical energy and apply the work-energy theorem.
First, let's find the potential energy of the elevator when it first contacts the spring. The potential energy is given by the formula:
Ep = mgh
where m is the mass of the elevator (2000 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height from the point where it first contacts the spring to the ground (which is the maximum compression of the spring, 2.00 m in this case).
Ep = (2000 kg)(9.8 m/s^2)(2.00 m) = 39,200 J
Next, let's find the work done on the elevator by the frictional force during the downward motion of 1.00 m. The work done is given by the formula:
Wf = Fd
where F is the force of friction (17,000 N) and d is the distance traveled (1.00 m).
Wf = (17,000 N)(1.00 m) = 17,000 J
Since the elevator is initially falling at 4.00 m/s, it has kinetic energy which needs to be subtracted from the potential energy and the work done by friction to find the final speed.
Now, let's find the initial kinetic energy of the elevator using the formula:
Ek = (1/2)mv^2
where m is the mass of the elevator (2000 kg) and v is the initial velocity (4.00 m/s).
Ek = (1/2)(2000 kg)(4.00 m/s)^2 = 16,000 J
Using the conservation of mechanical energy, we can write the equation:
Ep + Ek + Wf = Ef
where Ef is the final mechanical energy.
Because energy is conserved, the final mechanical energy (Ef) will be equal to the initial mechanical energy (Ep + Ek) minus the work done by friction (Wf).
Ef = (Ep + Ek) - Wf
Ef = (39,200 J + 16,000 J) - 17,000 J
Ef = 38,200 J
Finally, we can find the final velocity (vf) using the formula for kinetic energy:
Ef = (1/2)mvf^2
Rearranging the equation and solving for vf:
vf = √(2Ef / m)
vf = √[(2)(38,200 J) / (2000 kg)]
vf ≈ 7.76 m/s
Therefore, the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is approximately 7.76 m/s.