Figure shows a heavy block kept on a frictionless surface and being pulled by two ropes of equal mass m . At t=0, the force on the left rope is withdrawn but the force on the right end continues to act. Let F1 and F2 be the magnitudes of the forces by the right rope and the left rope on the block respectively. Then

Question

(A) F1=F2=F for t<0 (B) F1=F22=F+m g for t<0 (C) F1=F, F2=F for t>0 (D) F10

Tardigrade Admin · Accepted Answer

(a) is valid due to symmetry and since acceleration is 0 initially

F_{1} = F

and similarly

F_{2} = F

(b) is out, since

m g

is a vertical force and has got nothing to do with horizontal forces.
(c) and

(d)

are ruled out, as

F_{2} \neq = F

as

F

acts on the whole system and

F_{2}

acts on

m

only for

t > 0

$F_{1} = F_{2} = F$ for $t < 0$

$F_{1} = F_{2}^{2} = F + m g$ for $t < 0$

$F_{1} = F, F_{2} = F$ for $t > 0$

$F_{1} < F, F_{2} = F$ for $t > 0$

F1​=F2​=F for t<0

F1​=F22​=F+mg for t<0

F1​=F,F2​=F for t>0

F1​<F,F2​=F for t>0

(a) is valid due to symmetry and since acceleration is 0 initially F1​=F and similarly F2​=F (b) is out, since mg is a vertical force and has got nothing to do with horizontal forces. (c) and (d) are ruled out, as F2​=F as F acts on the whole system and F2​ acts on m only for t>0

$F_{1} = F_{2} = F$ for $t < 0$

$F_{1} = F_{2}^{2} = F + m g$ for $t < 0$

$F_{1} = F, F_{2} = F$ for $t > 0$

$F_{1} < F, F_{2} = F$ for $t > 0$