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Byju's Answer

Standard VIII

Physics

Free Body Diagram

Figure shows ...

Question

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on a frictionless surface and being pulled by two ropes of equal mass m. At $t = 0$ , the force on the left rope is withdrawn but the force on the right end continues to act.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 F $_{1}$ and F $_{2}$ be the magnitudes of the forces by the right rope and the left rope on the block respectively, then:

F_{1} = F_{2} = F

for

t < 0

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F_{1} = F_{2} = F + m g

for

t < 0

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F_{1} = F, F_{2} = F

for

t > 0

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F_{1} < F, F_{2} = F

for

t > 0

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Solution

The correct option is A $F_{1} = F_{2} = F$ for $t < 0$
For t < 0,
two forces were acting on the box, still it was at rest. This means that the forces were balanced.
Hence, as shown in the figure,
$F_{1} = F_{2}$
Since, these have to be equal to the forces shown in the figure,
$F_{1} = F_{2} = F$

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Similar questions

Q. A heavy block kept on a frictionless surface and being pulled by two ropes of equal mass

m

as shown in the figure. At

t = 0

, the force on the left rope is withdraw but the force on the right end continues to act, Let

F_{1}

and

F_{2}

be the magnitudes of the force by the right rope and the left rope on the block respectively.

Q. Two forces

F_{1}

and

F_{2}

act on a thin uniform elastic rod placed in space. Force

F_{1}

acts at right end of the rod and

F_{2}

acts exactly at centre of the rod as shown (both forces act parallel to length of the rod).

(i)

F_{1}

causes extension of rod while

F_{2}

causes compression of rod.

(i i)

F_{1}

causes extension of rod

F_{2}

also causes extension of rod.

(i i i)

F_{1}

causes extension of rod while

F_{2}

does not change length of rod.

The correct order of

T r u e / F a l s e

for above statements is

Q. Initially both blocks are at rest on a horizontal surface and string is just tight. At

t = 0

, two constant horizontal forces

F_{1}

and

F_{2}

start acting on blocks as shown.

f_{1}

and

f_{2}

are friction forces acting on 10 kg and 20 kg block (coefficient of friction between blocks and ground are 0.5). Values of

F_{1}

and

F_{2}

are given in column - I. Then match magnitudes of

f_{1}

f_{2}

and direction of

_{1}

with corresponding values of

F_{1}

and

F_{2}

given in column - II

[g = 10 m / s^{2}]

\begin{matrix} C o l u m n - I & C o l u m n - I I (A) & F_{2} = 120 N, F_{1} = 40 N & (p) & f_{2} = 100 N, f_{1} = 20 N (B) & F_{2} = 160 N, F_{1} = 40 N & (q) & f_{2} = 20 N, f_{1} = 50 N (C) & F_{2} = 60 N, F_{1} = 90 N & (r) & f_{2} = 70 N, f_{1} = 50 N (D) & F_{2} = 20 N, F_{1} = 90 N & (s) & u n i t v e c t o r i n d i r e c t i o n o f_{1} i s^i (t) & u n i t v e c t o r i n d i r e c t i o n o f_{1} i s -^i \end{matrix}

Two blocks of equal mass m are connected by an un-stretched spring and the system is kept at rest on α frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. If the extension of the spring is $x_{0}$ at time t, find the displacement of the two blocks at this instant.

Q. Let

f

be any continuous function on

[0, 2]

and twice differentiable on

(0, 2) .

f (0) = 0,

f (1) = 1

and

f (2) = 2,

then

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The correct option is A F1=F2=F for t<0For t < 0,two forces were acting on the box, still it was at rest. This means that the forces were balanced. Hence, as shown in the figure, F1=F2Since, these have to be equal to the forces shown in the figure,F1=F2=F

The correct option is A $F_{1} = F_{2} = F$ for $t < 0$
For t < 0,
two forces were acting on the box, still it was at rest. This means that the forces were balanced.
Hence, as shown in the figure,
$F_{1} = F_{2}$
Since, these have to be equal to the forces shown in the figure,
$F_{1} = F_{2} = F$