A 100 newton force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction is 0.5. If the accleration due to gravity is taken as 10m/s^2, the accleration of the block will be
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Fext=100N
m=10kg
coff of friction(u)=0.5
if R is the normal reaction by given by the surface
R=mg=10×10=100N
as frictional force acting on the block is
f=uR
f=umg
f=0.5×100
f=50N opposite to ext force
Net force on block
Fnet=Fext_f
accelration on the block
a=F_f/m
a=100_50/10
a=5m/s^2
m=10kg
coff of friction(u)=0.5
if R is the normal reaction by given by the surface
R=mg=10×10=100N
as frictional force acting on the block is
f=uR
f=umg
f=0.5×100
f=50N opposite to ext force
Net force on block
Fnet=Fext_f
accelration on the block
a=F_f/m
a=100_50/10
a=5m/s^2
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