Question

A 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same constant braking force, and are brought to rest. which object travels the greater distance before stopping? a 4.0-kg object is moving with speed 2.0 m/s. a 1.0-kg object is moving with speed 4.0 m/s. both objects encounter the same constant braking force, and are brought to rest. which object travels the greater distance before stopping? the 1.0-kg object the 4.0-kg object both objects travel the same distance. it is impossible to know without knowing how long each force acts.

Answer 1

Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
$F=ma$
from which we find an expression for the acceleration:
$a= \frac{F}{m}$ (1)

Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
$v_f^2 - v_i^2 = 2 a S$ (2)
where
$v_f$ is the final speed of the object
$v_i$ is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing $v_f$ (since the final velocity of the two objects is zero), we find
$-v_i^2 = 2 \frac{F}{m}S$
$S=- \frac{v_i^2 m}{2F}$
where we can ignore the negative sign (because the force F will bring another negative sign).

For the first object, we have
$S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m]$
And for the second object we have
$S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]$

And since the braking force applied to the two objects is the same, the two objects cover the same distance.