Respuesta :
Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
[tex]F=ma[/tex]
from which we find an expression for the acceleration:
[tex]a= \frac{F}{m} [/tex] (1)
Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
[tex]v_f^2 - v_i^2 = 2 a S[/tex] (2)
where
[tex]v_f[/tex] is the final speed of the object
[tex]v_i[/tex] is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing [tex]v_f[/tex] (since the final velocity of the two objects is zero), we find
[tex]-v_i^2 = 2 \frac{F}{m}S [/tex]
[tex]S=- \frac{v_i^2 m}{2F} [/tex]
where we can ignore the negative sign (because the force F will bring another negative sign).
For the first object, we have
[tex]S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m] [/tex]
And for the second object we have
[tex]S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m] [/tex]
And since the braking force applied to the two objects is the same, the two objects cover the same distance.
[tex]F=ma[/tex]
from which we find an expression for the acceleration:
[tex]a= \frac{F}{m} [/tex] (1)
Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
[tex]v_f^2 - v_i^2 = 2 a S[/tex] (2)
where
[tex]v_f[/tex] is the final speed of the object
[tex]v_i[/tex] is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing [tex]v_f[/tex] (since the final velocity of the two objects is zero), we find
[tex]-v_i^2 = 2 \frac{F}{m}S [/tex]
[tex]S=- \frac{v_i^2 m}{2F} [/tex]
where we can ignore the negative sign (because the force F will bring another negative sign).
For the first object, we have
[tex]S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m] [/tex]
And for the second object we have
[tex]S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m] [/tex]
And since the braking force applied to the two objects is the same, the two objects cover the same distance.
Since [tex]\rm s_1 \;and\;s_2[/tex] are same, therefore both the objects covers the same distance.
Given :
Mass of both the object -- [tex]\rm m_1 = 4\;Kg\;and\; m_2 = 1\;Kg[/tex]
Initial velocity of both the object -- [tex]\rm u_1 = 2\;m/sec \; and \; u_2 = 4\;m/sec[/tex]
Final velocity of both the object -- [tex]\rm v_1 = 2\;m/sec \; and \; v_2 = 4\;m/sec[/tex]
Solution :
We know that,
F = ma
[tex]\rm a= \dfrac{F}{m}[/tex] ---- (1)
We know that,
[tex]\rm v^2 = u^2+2as[/tex] --- (2)
Where,
final velocity - v,
initial velocity - u,
a is acceleration,
and s is displacement.
For object 1, from equation (1) and (2),
[tex]\rm s_1 = -\dfrac{u_1^2m_1}{2F}[/tex]
[tex]\rm s_1 = -\dfrac{2^2\times 4 }{2F} = -\dfrac{8}{F}[/tex]
For object 2, from equation (1) and (2),
[tex]\rm s_2 = -\dfrac{u_2^2m_2}{2F}[/tex]
[tex]\rm s_2 = -\dfrac{4^2 \times 1}{2F} = - \dfrac {8}{F}[/tex]
Since [tex]\rm s_1 \;and\;s_2[/tex] are same, therefore both the objects covers the same distance.
For more information, refer the link given below
https://brainly.com/question/11934397?referrer=searchResults
