Answer:
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping
Explanation:
As we know that both mass have same horizontal force opposite to their motion
So we will have
[tex]a = \frac{F}{m}[/tex]
[tex]a_1 = \frac{F}{4}[/tex]
[tex]a_2 = \frac{F}{2}[/tex]
now the stopping distance of an object moving with initial speed v is given as
[tex]v_f^2 - v_i^2 = 2(-a) d[/tex]
[tex]d = \frac{v^2}{2a}[/tex]
so here we have
[tex]d_1 = \frac{2^2}{\frac{F}{4}}[/tex]
[tex]d_1 = \frac{16}{F}[/tex]
for other object we have
[tex]d_2 = \frac{4^2}{\frac{F}{2}}[/tex]
[tex]d_2 = \frac{32}{F}[/tex]
So correct answer will be
B) The 2-kg mass travels twice as far as the 4-kg mass before stopping