Question

a 40 - kilogram mass is moving across a horizontal surface at 5.0 m/s. what is the magnitude of the net force required to bring the mass to a stop in 8.0 seconds? 25 n 40 n 1 n 5 n

Explanation:

Step1: Calculate the acceleration

Use the formula $a=\frac{v - u}{t}$, where $v = 0$ (final velocity, as the mass stops), $u=5.0$ m/s (initial velocity), and $t = 8.0$ s (time).
$a=\frac{0 - 5.0}{8.0}=-\frac{5.0}{8.0}=- 0.625$ m/s². The negative sign indicates deceleration.

Step2: Calculate the net - force

Use Newton's second - law $F = ma$, where $m = 40$ kg (mass) and $a=-0.625$ m/s².
$F=40\times0.625 = 25$ N (we take the magnitude, so we ignore the negative sign of acceleration for force magnitude calculation).

Answer:

25 N