A block of mass 5 kg is on a rough horizontal surface
and is at rest. Now a force of 24 n is imparted to it with
negligible impulse. If the coefficient of kinetic friction is
0.4 and g = 9.8m / s2
, then the acceleration of the block
is
Answers
Answered by
0
Answer:
Explanation:
0.88 m/s2
24-(0.4*9.8*5)=4.4
Acceleration= 4.4/5=0.88
Answered by
0
Answer:
a = 0.88 m/s2
Explanation:
given,
m = 5 kg
f =24 N
m ( coefficient of kinetic friction ) = 0.4
g= 9.8 m/ s2 , n is the normal force
net force ( ma) ,
= applied force - kinetic friction
= f - mn
= 24 - 0.4( 5×9.8)
= 24 - 19.6
net force = 4.4 N
m a = 4.4
a = 4.4/ m
a = 4.4/5
a = 0.88 m/ S2
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