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In the figure shown, F is in newton and ...

In the figure shown, `F` is in newton and `t` in seconds, Take `g =10 m//s^(2)`.
(a) Plot acceleration of the block versus time graph.
(b) Find force of friction at,`t =2_(s) and t =8_(s).`

Text Solution

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(a) Normal reation, `R =mg =20N`. Limiting value of friction ,`f_(1) =muR=0.6xx20= 12N` The applied force `F(=2t)` crosses this limiting value of friction at `6_(s)`, therefore ,upto `6_(s)` block remain stationary and after `6_(s)` it starts moving. After `6_(s)` friction becomes constant at `12N` but the applied force keeps on increasing. Therefore, acceleration keeps on increasing.
For `tle6_(s)`
`f =F=2t` ...(i)
`F_("net") =F-f =0`
`a = (F_("net"))/(m) =0`
For `tgt6_(s)`
`f = 2t`
`f = 12N=f_(1)` ...(ii)
`F_("net") =F-f =2t-12`
`a = (F_("net"))/(m) =(2t-12)/(2) =(t-6)`
`:. a-t` graph is a straight line with slop ` =1` and intercept ` =-6`. Corresponding `a-t` graph is as shown.
(b) At `t=2_(s),f =4N` [fromEq. (i)]
At `t=8_(s),f =12N` [fromEq. (ii)]
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