The correct options are
A velocity of box relative to platform after
4 s is
3 m/s.
D friction force between man and platform is
30 N.
The FBDs are as shown
![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/802508/original_download_287129.png)
From the FBD of man, we have
N=80g=800 N Maximum friction force between platform
A and man is
fmax=μN=0.5×800=400 N ∵ 400N>50N, there will be no slipping between platform and man.
So, taking man and platform as a system, the acceleration of the platform is given as
aA=50mM+mA=5080+120=0.25 m/s2 From the FBD of box B, we have
acceleration of the box as
aB=50mB=50100=0.5 m/s2 So, relative acceleration is
aBA=aB−aA=(0.5)−(−0.25)=0.75 m/s2 Thus, relative velocity after
t=4 s will be given by
vBA=uBA+aBAt ⇒ vBA=0+0.75×4=3 m/s Also, friction between man and platform
A is
fs=mA×aA=120×0.25=30 N