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Question: Two blocks of masses 10 kg and 20 kg are connected by a massless string and are placed on a smooth h...

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 :

A

100 N

B

200 N

C

300 N

D

400 N

Answer

400 N

Explanation

Solution

Here m1=10 kg,m2=20 kg,F=600 Nm _ { 1 } = 10 \mathrm {~kg} , m _ { 2 } = 20 \mathrm {~kg} , F = 600 \mathrm {~N}

Let T be tension of the string and a

be common acceleration of the system

∴a=Fm1+m2=600 N10 kg+20 kg=60030 ms−2=20 ms−2\therefore a = \frac { F } { m _ { 1 } + m _ { 2 } } = \frac { 600 \mathrm {~N} } { 10 \mathrm {~kg} + 20 \mathrm {~kg} } = \frac { 600 } { 30 } \mathrm {~ms} ^ { - 2 } = 20 \mathrm {~ms} ^ { - 2 }

when a force F is applied on 10 kg block, then the tension in the string is

T=m2a=(20 kg)(20 ms−2)=400 NT = m _ { 2 } a = ( 20 \mathrm {~kg} ) \left( 20 \mathrm {~ms} ^ { - 2 } \right) = 400 \mathrm {~N}