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A book of mass 5 kg is kept on a horizon...

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 ?

Text Solution

Verified by Experts

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.
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Knowledge Check

  • A block A of mass m_(1) rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m_(2) is suspended. The coefficient of kinetic friction between the block and the table is mu_(k) . When the block A is sliding on the table the tension in the string is

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