Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force F = 10 N,, then tension $T_{1} =$

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10N

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Solution

The correct option is C
8N

By comparing the above problem with general expression. $T_{1} = \frac{(m_{2} + m_{3})}{m_{1} + m_{2} + m_{2}} = \frac{(3 + 5) 10}{2 + 3 + 5} = 8 N e w t o n$

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Similar questions

Q. Three blocks of mass

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and

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are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force of

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, tension

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Q. Three blocks of masses

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are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force

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Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force F = 10 N,, then tension $T_{1} =$

Q. Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth horizontal surface as shown in the figure. If a force F=600 N is applied to 10 kg block, then the tension in the string is:

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Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force F = 10 N,, then tension T1=

The correct option is C 8N By comparing the above problem with general expression. T1=(m2+m3)m1+m2+m2=(3+5)102+3+5=8 Newton

Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force F = 10 N,, then tension $T_{1} =$

The correct option is C
8N

By comparing the above problem with general expression. $T_{1} = \frac{(m_{2} + m_{3})}{m_{1} + m_{2} + m_{2}} = \frac{(3 + 5) 10}{2 + 3 + 5} = 8 N e w t o n$