Amrit Raj
Last Activity: 7 Years ago
As we see in the fig. the block of mass ‘m’ is in an accelerated frame with the vechile with a=20m/s2.
So a pseudo force comes into existence which acts in the opposited direction of acceleration on the block of mass ‘m’ at he contact surface of the block and the cart.
And value of force F(Pseudo)=m.a=20.m ,in this case.
Now a equal amout of normal reaction also acts on the block at the contact surface.
So, F(Normal reaction)=20m
Applying newtons law on the block of mass ‘m’ in vertical direction,
we get,
f-mg=0
=>$N-mg=0 (where ,f=$N ,$=cofficient of friction,N=Normal Reaction)
=>20m.$=mg (N=20.m)
=>$=g/20
=>$=10/20=0.5(coefficient of friction)
