The masses of the blocks are Ma=6.00 kg, Mb=1.00 kg, Mc=4.00 kg, and there is fr
ID: 1311204 • Letter: T
Question
The masses of the blocks are Ma=6.00 kg, Mb=1.00 kg, Mc=4.00 kg, and there is friction between the horizontal plane and Ma, (?k?0). Ma is observed to travel at a constant velocity.
A.
Tx is .... Tz
The magnitude of the total force on Ma is .... 0
Ma is moving to the left
Mc accelerates downward
Tx is .... Ty
Mc*g is .... Ty
options: greater than, less than, equal, true ,false
B. Calculate the coefficient of kinetic friction between Ma and the horizontal surface.
C. Mass Mc is now increased by 0.80 kg. Calculate the magnitude of the acceleration of Ma.
Explanation / Answer
Tx=Tz
magnitude of total force=0
as Ma is travelling with constant velocity,the whole system is travelling with constant velocity.
then for Mb:
Tz=1*9.8=9.8 N (assuming the system is moving towards right)
Tx=Tz=9.8 N
Ty=4*9.8=39.2 N
so as Tw>Tx,our assumption is correct.
Ma is moving towards right.
Mc is moving down
Tx is lesser than Ty.
Mc*g is equal to Ty.
B.
let coeffcient of friction be k.
then for Ma,
(Ty-Tx)=6*9.8*k
k=0.5
C.
the net increase in force=0.8*9.8=7.84
so acceleration=7.84/(6+4+1)=0.712 m/s^2
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