1.

A 40N force pulls a system of three masses on a horizontal frictionless surfacethevalue of tension T_(1)is

Answer»

10 N
20 N
30 N
40 N

Solution :
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`


Discussion

No Comment Found