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A 40N force pulls a system of three masses on a horizontal frictionless surfacethevalue of tension T_(1)is |
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Answer» 10 N ACCELERATIONOF the system `"a"= ("Force" )/( "Totalmass" )` `= (40)/( 10 + 6 + 4)=(40)/(20) =2ms^(-2)` `F- T_(1)ma` `40 - T_(1) = 10 XX 2` `40 - T_(1)= 20` `:.T_(1)= 20 - 40 =20` `T_(1)= 20N` |
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