(a)KE1=m•v1²/2=24.6 J.
v2=v1/2
KE2= m•v2²/2 =
=m•v1²/2•4=KE1/4=
=24.6/4=6.15 J.
(b)W(fr)=KE2-KE1=
6.15-24.6 = -18.45 J.
W(fr)=F(fr) •s•cos180º=>
F(fr) = W(fr)/ s•cos180º=
=-18.45/3.2•(-1)=5.77 N.
(c) KE2=- W(fr1)= - F(fr) •s1•cos180º
s1 = KE2/ F(fr)•cos180º=
=6.15/(-5.77) •(-1)=1.07 m.
A block is initially sliding along a horizontal surface with 24.6 J of kinetic energy. Friction causes the block’s speed to be halved in a distance of 3.20 m.
a) What is the kinetic energy of the block after traveling the 3.20 m?
b) What is the magnitude of the kinetic frictional force that acts on the block?
c) How much further will the block travel before coming to a complete stop?
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