2. what is the force on a 1000 kg elevator that is falling freely at 9.8 m/sec²?

Explanation:

Step1: Identify the formula

Use Newton's second - law $F = ma$.

Step2: Substitute values

$m = 1000$ kg, $a=9.8$ m/s², so $F=1000\times9.8$.

Step3: Calculate the force

$F = 9800$ N.

Answer:

9800 N