Safety engineers estimate that an elevator can hold a maximum mass of 1000 kg (including the mass of the elevator). Tensile strength (force) tests show that the cable supporting the elevator can tolerate a maximum upward force of 25,000 N. What is the maximum acceleration that the elevator's motor can produce without breaking the cable?

Assuming that the elevator behaves as an isolated system, there are two forces acting on it at any time: the tension T in the cable (acting upward), and gravity, acting downward with a constant acceleration g = 9.8 m/s2 over the total mass (in this case 1000 kg).
So, according 2nd Newton's Law, F = m*a, we can write the following expression:
F = T - m*g (1)

⇒ [tex]T- m*g = m*a (2)[/tex]

The maximum force, will be applied when T reaches to its maximum possible value, 25,000 N.
Replacing the values of Tmax, m and g in (2) we can solve for a, as follows:

[tex]a = \frac{T-m*g}{m} = \frac{25,000N- 1000kg*9.8m/s2}{1000kg} = 15.2 m/s2 (3)[/tex]

So, the maximum acceleration possible without breaking the cable is 15.2 m/s2.