A book of mass 5 kg is kept on a horizontal table. It is connected to a weight of mass 2 kg by a weightless string passing over a smooth pulley . The part of the string on the table is horizontal, and the weight is hanging freely from the pulley. If the coefficient of kinetic friction between the table and the block is 0.2, find the acceleration of the block. What will be the tension in the string ?

Explanation

Weight of the block, Mg = `5xx9.8` N,
`:.` Normal force of the table on the block,
R = `5xx 9.8 `N
Kinetic friction against the motion of the block,
f.= `mu R=0.2xx5xx9.8` = 9.8N
Let the tension in the string be T and acceleration of the block be a fig.
Hence for the motion of the block, T-f. = Ma
or, T-9.8= 5a `" "cdots(1)`
The weight of mass m (say ) moves downwards with the same acceleration.
`:.` mg-T = ma
or, `2xx9.8-T = 2a [ becausem=2] " "cdots(2)`
Adding (1) and (2) we get,
`2xx9.8`-9.8=7a
or, a=`(9.8)/(7)=1.4 "m.s"^(-2)`
Substituting this value in (1), we get,
T-9.8 = `5xx1.4 " "`or, T =16.8 N.