A block of mass 5 kg is pulled by a force F as shown in the figure. If the coefficient of friction is 0.1, then te force needed to accelerate the block to `3 m//s^(2)` to the right is close to

Options

A. 12N
B. 22N
C. 32N
D. 42N

Answer

B

Explanation

Component of force accelerating the block is `F cos theta`.
Net force on block `F_("net") = F cos theta - mu N`
`=F cos 30^(@) - 0.1 xx 5 xx 9.8`
`=(sqrt3F)/(2) - 4.9`
From `F_("net") = ma`, we have,
`(sqrt3)/(2) F-4.9 = 5 xx 3 rArr F = (19.9)/(sqrt3) = 22.9N`