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A 42 kg slab rests on a frictionless flo...

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)`)

Text Solution

Verified by Experts


(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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