# an observer and a vehicle both start moving together from rest with acc 5 m/s^2 and 2 m/s^2 in same directions .there is a 2 kg block on the floor of the vehicle and coeff of friction is 0.3 betwwen their surfaces . find the work done by the  frictional force on the block as observed by the running observer during first 2 secinds of motion ?Pls reply soon thnx

Chetan Mandayam Nayakar
312 Points
10 years ago

limiting friction=kmg=2*10*0.3=6N, inertial force=2*2=4N, so there is no slipping. Force of friction on block=4N, distance travelled during first two seconds of motion=(1/2)*3*4=6m, therefore work done=6m*4N =24 Joule

Swamynathanvp
14 Points
2 years ago
fmax = 0.3×2×10
a of block wrt ground is 2m/s2 which is provided by f = ma=2×2=4N

Relative a = 2-5= -3m/s2
Displacement=1/2at2= 1/2×-3×2×2=-6m

Work done = f.s = 4 ×-6=-24J

relkjew
15 Points
one year ago
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