Figure shows a heavy block kept on a frictionless surfaces and being mass m. At `t=0`, the force on the left rop is withdrawn but the force on the right end continues to act. Let `F_(1) and F_(2)` be the magnitudes of the forces acting on the block by the right rope and the left rope on the block respectively, then :

Options

A. (a)`F_(1) = F_(2) = F` for `t lt 0`
B. (b)`F_(1) = F_(2) = F + mg` for `t lt 0`
C. (c)`F_(1) = F, F_(2) = F` for `t gt 0`
D. (d)`F_(1) lt F, F_(2) lt F` for `t gt 0`

Answer

A, D

Explanation

For t < 0
`implies F = F_(1) = F_(1)`
For t > 0
`F_(1) - F_(2) = Ma`……M is mass of block
`F - F_(1) = ma`
`F_(2) = ma implies F gt F_(1) gt F_(2)`