Question

A man sits on a chair supported by a rope passing over a friction less fixed pulley .the man who weighs 1000n exerts a force of 450n on the chair downwards whule pulling the rope on the other side . if the chair weighs 250n tgen acceleration of chair is?

Answer 1

If we take man + chair as a system then for the whole system net force is given by

$F_{net} = 2T - (m + M)g$

$F_{net} = 2T - 1000 - 250$

as per Newton's II law

Net force on the system is product of mass and acceleration

$F_{net} = (m + M)a$

$(100 + 25)a = 2T - 1250$

$125a = 2T - 1250$

now again for the net force on chair

$F_{chair} = T - mg - N$

$F_{chair} = T - 250 - 450$

again by newton's 2nd law we can say

$m_{chair}*a =F_{chair}$

$25*a = T - 700$

now put the value of T from 2nd equation to 1st equation

$T = 25*a + 700$

$125a = 2*(25a + 700) - 1250$

$125a - 50a = 1400 - 1250$

$75a = 150$

$a = 2m/s^2$

so the acceleration of the whole system will be 2m/s^2 upwards

Answer 2

Answer:

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