Final answer:
Using the conservation of energy principle, the work done by gravity on the elevator is 180,427.5 J, the velocity of the elevator just before striking the spring is 26.18 m/s, and the amount the spring compresses is 1.93 m.
Step-by-step explanation:
To solve the problem involving the elevator and the spring, we need to use the conservation of energy principle.
Work Done by Gravity on the Elevator
The work done by gravity (Wg) on the elevator before it hits the spring is equal to the change in gravitational potential energy, which can be calculated using the formula:
Wg = m × g × h
where m is the mass of the elevator (525 kg), g is the acceleration due to gravity (9.81 m/s2), and h is the height (35 m).
Wg = 525 kg × 9.81 m/s2 × 35 m = 180,427.5 J
Velocity of the Elevator Before Striking the Spring
The velocity (v) of the elevator just before striking the spring can be found using the energy conservation equation:
0.5 × m × v2 = m × g × h
Solving for v gives:
v = √(2 × g × h) = √(2 × 9.81 m/s2 × 35 m) = 26.18 m/s
Compression of the Spring
The amount the spring compresses (x) can be found by equating the work done by the spring (which is also the potential energy stored in the spring when fully compressed) to the kinetic energy of the elevator just before impact:
0.5 × k × x2 = 0.5 × m × v2
0.5 × 9.60×104 N/m × x2 = 180,427.5 J
Solving for x gives:
x = √(2 × 180,427.5 J / 9.60×104 N/m) = 1.93 m