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A 5 kg block is kept on a horizontal platform at rest . At t=0 , the platform starts moving with a constant acceleration of 1 m/s^2 . The coefficient of friction between block and the platform is 0.2 . The work done force of friction on the block in reference frame fixed with ground in 10 sec is

In the ques.. if coefficient of friction becomes 0.02 , the work done . (same conditions applied)

from work energy theorem the work done by all forces is equal to change in kinetic energy

as initially the block is at rest wrt ground

${W}_{normalforce}+{W}_{friction}+{W}_{mg}={K}_{final}-{K}_{initial}={K}_{final}(as{K}_{initial}=0)\phantom{\rule{0ex}{0ex}}{W}_{friction}={K}_{final}(as{W}_{normalforce}and{W}_{mg}=0\mathrm{sin}ceforceanddiplacementareperpendicular)....1\phantom{\rule{0ex}{0ex}}\frac{1}{2}\times 5\times (velocityafter10second{)}^{2}=\frac{1}{2}\times 5\times (0+1\times 10{)}^{2}=250J\phantom{\rule{0ex}{0ex}}belowcompletlydescribeswhyblockisatrestwrtplatformandhassameacclerationasthatofplatform\phantom{\rule{0ex}{0ex}}themaximum\left(limiting\right)frictionforwhichtheblockhasnorelativemotionwrtplatform=\mu \times massofblock\times g=0.2\times 5\times 10=10N\phantom{\rule{0ex}{0ex}}nowduetoacclerationofplatformfrictionforceappliedbyplatformonblockisf=ma=5\times 1=5N(aslockhasnorelativemotionwrtplatform)\phantom{\rule{0ex}{0ex}}henceweknowthatfrictionisselfadjustinginnaturesofrictionforcewillbe5N.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

but if coefficient of friction has value 0.002 than max friction is 1N (same as above ) and due to motion of platform

5N force is required (as from above) which makes the body to give some acceleration wrt platform

$nownetforceonblockwrtground=(5-1)backward=4N\phantom{\rule{0ex}{0ex}}a=\frac{4}{5}(backwarddirection)\phantom{\rule{0ex}{0ex}}velocityafter10second=0+\frac{4}{5}\times 10=8m/s(u\mathrm{sin}gv=u+at)\phantom{\rule{0ex}{0ex}}nowworkdonebyfrictionis=\frac{1}{2}\times 5\times (8{)}^{2}=160J$

Regards

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