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Byju's Answer

Standard IX

Physics

Second Law of Motion

As shown in t...

Question

As shown in the figure, three masses $m, 3 m, 5 m$ connected together lie on a frictionless horizontal surface and pulled to the left by a force $F$ . The tension $T_{1}$ in the first string is $24 N$ . Find acceleration of the system (ii) tension in the second string and (iii) force F.

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Solution

Total acceleration of the system will be,

$a = \frac{F}{m + 3 m + 5 m}$

Tension $T_{1}$ in the string is,

$24 N = 8 m \times \frac{F}{9 m}$

$F = 27 N$

Tension $T_{2}$ in the string is,

$T_{2} = 5 m \times \frac{27}{9 m}$

$= 15 N$

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As shown in the figure, three masses m,3m,5m connected together lie on a frictionless horizontal surface and pulled to the left by a force F . The tension T1 in the first string is 24N . Find acceleration of the system (ii) tension in the second string and (iii) force F.

Total acceleration of the system will be, a=Fm+3m+5m Tension T1 in the string is, 24N=8m×F9m F=27N Tension T2 in the string is, T2=5m×279m =15N

As shown in the figure, three masses $m, 3 m, 5 m$ connected together lie on a frictionless horizontal surface and pulled to the left by a force $F$ . The tension $T_{1}$ in the first string is $24 N$ . Find acceleration of the system (ii) tension in the second string and (iii) force F.

Total acceleration of the system will be,

$a = \frac{F}{m + 3 m + 5 m}$

Tension $T_{1}$ in the string is,

$24 N = 8 m \times \frac{F}{9 m}$

$F = 27 N$

Tension $T_{2}$ in the string is,

$T_{2} = 5 m \times \frac{27}{9 m}$

$= 15 N$