A 2000-kg elevator is at rest when its cable breaks. The elevator falls 26 m before it encounters a giant spring at the bottom o
f the elevator shaft. As the elevator falls, its safety clamp applies a constant frictional force of 17,000 N. 1. List all the forces acting on the elevator after the cable breaks but before it hits the spring. Determine the work done by each force on the elevator. Make sure these work terms have the appropriate signs (positive or negative). 2. Calculate the net work done on the elevator. 3. Use the work-kinetic-energy theorem to determine the speed of the elevator right before it encounters the spring. 4. After the elevator encounters the spring, the spring compresses by 0.6 m before the elevator comes to rest. The safety clamp continues to apply a constant 17,000 N force. List all the forces acting on the elevator while it falls this final 0.6 m. 5. For each force listed in question F4, determine an expression for the work it does on the elevator. Make sure each work term has the appropriate sign (positive or negative). 6. Use the work-kinetic-energy theorem to determine the value of the spring constant k.
The answer is A. Force is equal to mass x acceleration. The acceleration due to gravity is a constant so multiplying the mass by 2 would double the force.
