In the arrangement shown in figure, coefficient of friction between the two blocks is u = 1/2. The force of
friction acting between the two blocks in newton is 2k, find value of k
Im F= 2N
2kg)
F2 = 20N
4kg
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The maximum force of friction that can occur between the two blocks is
f
max
=μmg =
2
1
×2×10 = 10N
The acceleration of the system is
sumofthemasses
sumoftheforces
=
4+2
20+(−2)
=3ms
−2
The system of blocks has an acceleration in the left side. The force of friction opposing the motion of the 2kg block in the left side is ma + force acting in the direction of friction = (2kg×3ms
−2
)+2N =8N. Thus the force of friction acting between the two blocks is 8N.
solution
f
max
=μmg =
2
1
×2×10 = 10N
The acceleration of the system is
sumofthemasses
sumoftheforces
=
4+2
20+(−2)
=3ms
−2
The system of blocks has an acceleration in the left side. The force of friction opposing the motion of the 2kg block in the left side is ma + force acting in the direction of friction = (2kg×3ms
−2
)+2N =8N. Thus the force of friction acting between the two blocks is 8N.
solution
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