Block B of mass 2kg rests on block A of mass 10kg . All surfaces are rough with the value of coefficient of friction as shown in the figure. Find the minimum force F that should be applied on block A to cause relative motion between A and B. (g = 10m/s^2)
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MasterAryan:
how did (1764)^1/2 come, the whole process?And in the options 42 is not there.
The options are (a)24N (b)30N (c)48N (d)60N
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Mass of Block A
m1 = 10 Kg
and
Mas of Block B
m2 = 2 Kg
Thus total Mass m = m1 + m2
m = 10 + 2
m = 12 Kg
Thus
Force = m * g
= 12 * 10
= 120 N
m1 = 10 Kg
and
Mas of Block B
m2 = 2 Kg
Thus total Mass m = m1 + m2
m = 10 + 2
m = 12 Kg
Thus
Force = m * g
= 12 * 10
= 120 N
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