A force F is applied to the initially, stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 s is

Options

A. 4
B. `0.67`
C. `3.75`
D. `2`

Answer

B

Explanation

Here
`F = 5t^(2)`
`f_("max") = mu N = 5N`
`a = (5t^(2) - 5)/(10)`
`underset(0)overset(v)(int) dv = underset(1)overset(2)(int) (1)/(2) (t^(2) - 1) dt`
`v = (1)/(2)[(t^(3))/(3) - t]_(1)^(2)`
`v = (2)/(3) m//s`