When a box of mass 45kg, is given an initial speed of 5ms-1 it slides along a horizontal floor a distance of 3m before coming to a rest. what is the coefficient of kinetic friction between the box and the floor? (G = 10ms-2)
\( \frac{5}{6} \)
\( \frac{5}{12} \)
\( \frac{1}{3} \)
\( \frac{2}{3} \)
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Explanation:
Coefficient of friction=F/R
F=m×a
v²=u²+2as
0=5²+2×a×3
25=6a
a=4.167
F=45×4.167
F=187.515
R=45×10
R=450
Coefficient of friction=187.515/450=5/12=
0.4167
coefficient of friction=F/R
F=ma
v²=u²+2as
0=5²+2×a×3
25=6a
a=25/6
a=4.167
F=45×4.167
F=187.5
R=45×10=450
coefficient of friction=187.5/450=0.4167
5/12=0=4167
option B
Explanation:
Coefficient of friction=F/R
F=m×a
v²=u²+2as
0=5²+2×a×3
-25=6a
a=-4.167(deceleration) i.e=4.167
F=45×4.167
F=187.515
R=45×10
R=450
Coefficient of friction=187.515/450=5/12=
0.4167
Coefficient of friction=F/R
F=m×a
v²=u²+2as
0=5²+2×a×3
25=6a
a=4.167
F=45×4.167
F=187.515
R=45×10
R=450
Coefficient of friction=187.515/450=5/12=
0.4167
Explanation:
Coefficient of friction=F/R
F=m×a
v²=u²+2as
0=5²+2×a×3
-25=6a
a=-4.167(deceleration) i.e a=4.167
F=45×4.167
F=187.515
R=45×10
R=450
Coefficient of friction=187.515/450=5/12=
0.4167
