Force acting on a block versus time graph is as shown in figure Choose the correct options.`(g = 10 m//s^(-2))`

Options

A. At `t = 2 s`, force of friction is `2 N`
B. At `t = 8 s`, force of friction is `6 N`
C. At `t = 10 s`, acceleration of block is `2 ms^(-2)`
D. At `t = 12 s`, velocity of block is `8 ms^(-1)`

Answer

A, B, C, D

Explanation

`f_(max) = 0.3 xx 2 xx 10 = 6 N`
At `t = 2 s, F = 2 N lt f_(max)`
`:. f = F = 2 N`
At `t = 8 s, F = 8 N gt f_(max)`
`:. f = 6 N`
At `t = 10 s, F = 10 N gt f_(max)`
`:. f = 6 N`
`a = (F - f)/(m) = (10 - 6)/(2) = 2 m//s^(2)`
`F = f_(max) = 6 N at6 s`
For `6 s le t le 10 s`
`a = (F - f)/(m) = (t - 6)/(2) = 0.5t - 3`
`int _(0)^(v) dv = int adt = int _(6)^(10) (0.5t - 3) dt`
` v = 4 m//s`
`After 10 s`
`a = (F - f)/(m) = (10 - 6)/(2) = 2 m//s^(2)`
`= constant`
`:. v' = v + at`
`= 4 + 2 (12 - 10)`
`= 8 m//s`