If the surface shown in the figure is frictionless- and T 1-6 -3 N- find F-A- 120 NB- 100 N- 120 -3 ND- 100 -3

The correct option is A 120 NFBD of the 3 kg body; nbsp; nbsp;T_1 =3a a=T_1/3=6√(3)/3 =2√(3) m/s^2 Consider all the blocks as a system. FBD of the syste ...

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Byju's Answer

Standard XII

Physics

Tension in a String

If the surfac...

Question

If the surface shown in the figure is frictionless, and $T_{1} = 6 \sqrt{3} N$ , find $F$ .

120 N

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100 N

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120 \sqrt{3} N

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100 \sqrt{3}

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Solution

The correct option is A $120 N$
FBD of the $3 k g$ body;
$T_{1} = 3 a a = \frac{T_{1}}{3} = \frac{6 \sqrt{3}}{3} = 2 \sqrt{3} m / s^{2}$
Consider all the blocks as a system.
FBD of the system will be,

$m a = F c o s 30^{o}$
$30 \times 2 \sqrt{3} = F \times \frac{\sqrt{3}}{2}$
$F = 120 N$

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If the surface shown in the figure is frictionless, and T1=6√3 N, find F.

The correct option is A 120 NFBD of the 3 kg body; T1=3aa=T13=6√33=2√3 m/s2 Consider all the blocks as a system. FBD of the system will be, ma=F cos 30o 30×2√3=F×√32 F=120 N

If the surface shown in the figure is frictionless, and $T_{1} = 6 \sqrt{3} N$ , find $F$ .

The correct option is A $120 N$
FBD of the $3 k g$ body;
$T_{1} = 3 a a = \frac{T_{1}}{3} = \frac{6 \sqrt{3}}{3} = 2 \sqrt{3} m / s^{2}$
Consider all the blocks as a system.
FBD of the system will be,

$m a = F c o s 30^{o}$
$30 \times 2 \sqrt{3} = F \times \frac{\sqrt{3}}{2}$
$F = 120 N$