Question

The force acting on the block of mass `1 kg` is given by `F=5-2t`. The frictional force acting on the block after time `t=2` seconds will be: `(mu=0.2)`

A. `2N`
B. `3N`
C. `1N`
D. zero

Answer 1

Correct Answer - A

`f_("max")=10xx0.2=2N`
Initial force`=5N gt 2N`
`:.` block will move with acceleration
`a=(5-2t-f_(max))/(1)=5-2t-2`
`(dv)/(dt)=3-2t` , `v=3t-t^(2)`
(`because` at `t=0, v=0`) `v=0 rArrt=0, 3 sec`
`:.` at `t=2 sec` block is moving
`:. f_("max")` will act, i.e., frictional force acting `=2N`