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Question Number 20501 by Tinkutara last updated on 27/Aug/17

The force acting on the block is given by F = 5 − 2t. The frictional force acting on the block at t = 2 s. (The block is at rest at t = 0)

$$\mathrm{The}\:\mathrm{force}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{the}\:\mathrm{block}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${F}\:=\:\mathrm{5}\:−\:\mathrm{2}{t}.\:\mathrm{The}\:\mathrm{frictional}\:\mathrm{force}\:\mathrm{acting} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{block}\:\mathrm{at}\:{t}\:=\:\mathrm{2}\:\mathrm{s}.\:\left(\mathrm{The}\:\mathrm{block}\:\mathrm{is}\:\mathrm{at}\right. \\ $$$$\left.\mathrm{rest}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\right) \\ $$

Commented by Tinkutara last updated on 27/Aug/17

Commented by ajfour last updated on 27/Aug/17

J=∫_0 ^( t) [(5−2t)−f ]dt J=5t−t^2 −2t =3t−t^2 J=0 when t=0 and t=3s therefore block is moving until t=3s ; friction at t=3s, is equal to f=μmg =0.2×1×10 = 2N .

$${J}=\int_{\mathrm{0}} ^{\:\:{t}} \left[\left(\mathrm{5}−\mathrm{2}{t}\right)−{f}\:\right]{dt} \\ $$$${J}=\mathrm{5}{t}−{t}^{\mathrm{2}} −\mathrm{2}{t}\:=\mathrm{3}{t}−{t}^{\mathrm{2}} \\ $$$${J}=\mathrm{0}\:{when}\:{t}=\mathrm{0}\:{and}\:{t}=\mathrm{3}{s} \\ $$$${therefore}\:{block}\:{is}\:{moving}\:{until} \\ $$$${t}=\mathrm{3}{s}\:;\:{friction}\:{at}\:{t}=\mathrm{3}{s}, \\ $$$$\:\:\:{is}\:{equal}\:{to}\:{f}=\mu{mg}\:=\mathrm{0}.\mathrm{2}×\mathrm{1}×\mathrm{10} \\ $$$$\:\:=\:\mathrm{2}{N}\:. \\ $$

Commented by Tinkutara last updated on 27/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by Tinkutara last updated on 27/Aug/17

And another thing can you explain that maximum frictional force is μmg = 2 N and at t = 0, applied force = 5 N, so how the block remains at rest? How 2 N stops 5 N?

$$\mathrm{And}\:\mathrm{another}\:\mathrm{thing}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain} \\ $$$$\mathrm{that}\:\mathrm{maximum}\:\mathrm{frictional}\:\mathrm{force}\:\mathrm{is}\:\mu{mg} \\ $$$$=\:\mathrm{2}\:\mathrm{N}\:\mathrm{and}\:\mathrm{at}\:{t}\:=\:\mathrm{0},\:\mathrm{applied}\:\mathrm{force}\:=\:\mathrm{5}\:\mathrm{N}, \\ $$$$\mathrm{so}\:\mathrm{how}\:\mathrm{the}\:\mathrm{block}\:\mathrm{remains}\:\mathrm{at}\:\mathrm{rest}?\:\mathrm{How} \\ $$$$\mathrm{2}\:\mathrm{N}\:\mathrm{stops}\:\mathrm{5}\:\mathrm{N}? \\ $$

Commented by ajfour last updated on 27/Aug/17

at t=0 block was at rest, (given) even a little after block is moving friction tries to resist, and even external force decreases as time proceeds.the block gains speed until F=5−2t = μmg=2 that is till t=1.5s . After this friction is more =2, (←); block slows down, force keeps decreasing and equals zero at t=2.5s and thereafter negative, block stops at t=3s again, rests for a little time (meanwhile force changes and friction reverses direction, block then moves left, friction acts rightwards..

$${at}\:{t}=\mathrm{0}\:{block}\:{was}\:{at}\:{rest},\:\left({given}\right) \\ $$$${even}\:{a}\:{little}\:{after}\:{block}\:{is}\:{moving} \\ $$$${friction}\:{tries}\:{to}\:{resist},\:{and}\:{even} \\ $$$${external}\:{force}\:{decreases}\:{as}\:{time} \\ $$$${proceeds}.{the}\:{block}\:{gains}\:{speed} \\ $$$${until}\:{F}=\mathrm{5}−\mathrm{2}{t}\:=\:\mu{mg}=\mathrm{2} \\ $$$${that}\:{is}\:{till}\:{t}=\mathrm{1}.\mathrm{5}{s}\:.\:{After}\:{this} \\ $$$${friction}\:{is}\:{more}\:=\mathrm{2},\:\left(\leftarrow\right);\:{block} \\ $$$${slows}\:{down},\:{force}\:{keeps}\:{decreasing} \\ $$$$\:{and}\:{equals}\:{zero}\:{at}\:{t}=\mathrm{2}.\mathrm{5}{s} \\ $$$${and}\:{thereafter}\:{negative}, \\ $$$${block}\:{stops}\:{at}\:{t}=\mathrm{3}{s} \\ $$$${again},\:{rests}\:{for}\:{a}\:{little}\:{time} \\ $$$$\left({meanwhile}\:{force}\:{changes}\:{and}\right. \\ $$$${friction}\:{reverses}\:{direction},\: \\ $$$${block}\:{then}\:{moves}\:{left},\: \\ $$$${friction}\:{acts}\:{rightwards}.. \\ $$

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