The `10 kg` block is resting on the horizontal surface when the force `F` is applied to it for `7 s` . The variation of `F` with time is shown . Calculate the maximum velocity reached by the block and the total time `t` during which the block is in motion. The coefficient of static and kinetic friction are both `0.50.(g = 9.8 m//s^(2))`

Answer

B

Explanation

Block will start moving at , `F = mu mg`
or `25t = (0.5) (10) (9.8) = 49 N`
`:. t = 1.96 s`
Velocity is maximum at the end of `4 second`
`:. (dv)/(dt) = (25t - 49)/(10) = 2.5t - 4.9`
`:. int_(0)^(nu_max) dv = int_(1.96)^(4) (2.5 t - 49) dt`
`:. nu_(max) = 5.2 m//s`
For `4s lt t lt 7 s`
Net retardation ` a_(1) = (49 - 40)/(10) = 0.9 m//s^(2)`
`:. nu = nu_(max) - a(1) t_(1) = 5.2 - 0.9 xx 3 = 2.5 m//s`
For` t gt 7 s`
Retardation ` a_(2) = (49)/(10) = 4.9 m//s^(2)`
`:. t = (nu)/(a_(2)) = (2.5)/(4.9) = 0.51 s`
`:.` Total time `= (4 - 1.96) + (7 - 4) + (0.51)`
`= 5.55 s`