A 5 kg block is kept on a horizontal platform at rest At t=0 , the platform starts - Physics - Work Energy And Power

Question

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)

Rajnish Kumar · Accepted Answer

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
Regards