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A 42 kg slab rests on a frictionless floor A 9.7 kg block rests on the top of the slab as shown in the figure. The co-efficient of static friction between the block and the slab is 0.53, while the co-efficient of kinetic friction is 0.38. The 9.7 kg block is acted upon by a horizontal force of 110 N. What are the resulting accelerations of (a) the block, and (b) the slab? (Take g=9.8m//s^(2)) |
Answer» SOLUTION : (a) From FBD of m `F-mu_(k)mg=ma_(2)` `a_(2)=(F)/(m)-mu_(k)g=((110N))/((9.7kg))-(0.38)(9.8m//s^(2))` `=7.6m//s^(2)` (towards left) (B) From FBD of M `f=Ma_(1) rArr (mu_(k)mg)/(M)=a_(1)` `a_(1)=((0.38)(9.7kg)(9.8m//s^(2)))/((42kg))` `=0.86m//s^(2)` (towards left) |
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