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)

Dear student
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
Wnormal force+Wfriction+Wmg =Kfinal- Kinitial=Kfinal    (asKinitial =0)Wfriction =Kfinal   (as Wnormal force and Wmg =0 since force and diplacement are perpendicular)....1>                   12×5×(velocity after 10 second)2  =12×5×(0+1×10)2=250Jbelow completly describes why block is at rest wrt platform and has same accleration as that of platform the maximum(limiting) friction for which the block has no relative motion wrt platform=μ×mass of block×g=0.2×5×10=10Nnow due to accleration of platform friction force applied by platform on block is f =ma= 5×1=5N  (as  lock has no relative motion wrt platform)hence we know that friction is self adjusting in nature so friction force will be 5N .
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
now net force on block wrt ground = (5-1)backward=4Na = 45 (backward direction)velocity after 10 second= 0+45×10=8m/s   (using v=u+at)now work done by friction is =12×5×(8)2=160J


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It is 250 J.
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Approx 249J
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