Respuesta :
Acceleration is constant, so we can use the following kinematic equation and Newton's second law.
[tex]{v_f}^2 - {v_i}^2 = 2a\Delta x \implies -{v_i}^2 = -2\dfrac{F}{m}\Delta x \implies {v_i}^2 = 2\dfrac{F}{m} \Delta x[/tex]
where [tex]v_i[/tex] and [tex]v_f[/tex] are initial/final velocities, [tex]a[/tex] is acceleration, [tex]\Delta x[/tex] is displacment, [tex]F[/tex] is force, and [tex]m[/tex] is mass. Since the objects are coming to rest, the acceleration opposes the direction of motion and is negative.
Solve for [tex]\Delta x[/tex].
[tex]\Delta x = -\dfrac{m {v_i}^2}{2F}[/tex]
Compute the displacements of both objects.
[tex]\Delta x_{5.0\text{-kg}} = -\dfrac{(5.0\,\mathrm{kg}) \left(2.0\frac{\rm m}{\rm s}\right)^2}{2F} = -\dfrac{10}F \dfrac{\rm m}{\rm N}[/tex]
[tex]\Delta x_{1.25\text{-kg}} = -\dfrac{(1.25\,\mathrm{kg}) \left(4.0\frac{\rm m}{\rm s}\right)^2}{2F} = -\dfrac{10}F \dfrac{\rm m}{\rm N}[/tex]
Then both objects are displaced by the same amount.
Answer:
They stop in the same distance
Explanation:
f friction is the same for both and must overcome the KE of each object to bring each to rest.
f friction * d = 1/2 mv^2 for each object :
f * d1 = 1/2 m1 v1^2 f * d2 = 1/2 m2 v2^2
= 1/2 5 (2 )^2 = 1/2 1.25 (4)^2
Object 1 KE = 10 j Object 2 KE = 10 j
f d1 / f d2 = 10 / 10 ( f is the same for both)
d1 / d2 = 1 so d1 = d2 they stop in the same distance !
