A block of mass 2.00 kg slides across a rough floor with an initial speed
of 6.00 m/s. It comes to rest after sliding a distance Dx in time
to.
The coefficient of kinetic friction is 0.380.
(a) Calculate the value of the frictional force slowing down the block.
Draw a free body diagram for the block.
f
=
mkn
=
mkmg
=
(0.380)(2.00 kg)(9.80 N/kg)
=
7.45 N
(b) How far does the block slide before coming to rest?
a=-f/m=-7.45 N/2.00 kg=-3.72 m/s2
v2f
=
v2i+2aDx
0
=
(6.00 m/s)2+2(-3.72 m/s2)Dx
Dx
=
4.84 m
(c) How long does it take for the block to stop?
vf
=
vi+ato
0
=
6.00 m/s-(3.72 m/s2)to
to
=
1.61 s
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