A 6Kg block by a distance of s=12m along a horizontal surface at a constant speed. If the coefficient of kinetic friction is mu=0.2 and the cord pulling the block is at an angle of 45 degree with the horizontal, then calculate the work done by the person. (take g=10m/s2)
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Answer:
Given mass(m)= 5 kg; displacement (s)=20 m;μ
k
=0.20; θ=45
Normal force (vertical component)=F sin45
0
Applied Force =F cos45
0
Net forceF
net
=mgFsin45
o
C
Frictional force f=μ(mgFsin45)
o
C
Fcos45
0
=μ(mgFsin45
0
)
Fcos45
0
+μFsin45
0
=μmg
Fcos45
0
=
1+μtan45
0
μmg
.................(1)
Also we know work done =Force× displacement
=(
1+μ
μmg
)×s
=(
1.2
0.2×5×9.81
)×20
=163.5J
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