Consider a system of two masses and a pulley shown in the figure. The coefficient of friction between the two blocks and also between block and table is 0.1. Find the force F, that must be given to the 0.8 kg block such that it attains acceleration of `5 m//s^(2)`.
(Assume, acceleration due to gravity, `g=10 m//s^(2)`)

Options

A. 6.4 N
B. 7.1 N
C. 6.0 N
D. 7.8 N

Answer

A

Explanation

Given, acceleration of block, `a=5 m//s^(2)` and coefficient of friction between two blocks and table, `mu =0.1`

Free body diagram for 0.8 kg block.

For 0.8 kg block, friction force due to 0.2 kg block is in opposite direction of applied force F.
`:.` Equation of motion for 0.8 kg block,
`F-T-mu R-mu R_(1)=0.8 a`
Putting the given values, we get
`F-T-0.1 (0.8+0.2)g-0.1xx0.2 g=0.8 a`
`F-T-0.1xx10-0.1xx0.2xx10=0.8xx5`
`F-T=5.2` ...(i)
Free body diagram for 0.2 kg body,

For 0.2 kg block is in opposite direction of tension force.
So, `T-mu R_(1)=0.2a`
`T-0.1xx0.2g=0.2xx5`
`T-0.02xx10=1.0`
`T=1.2` ...(ii)
From Eqs. (i) and (ii), we get
`F-1.2=5.2`
`F=5.2+1.2=6.4 N`