A man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weighs `1,000N` exerts a force of `450N` on the chair downwards while pulling the rope on the other side. If the chair weighs `250N` then the acceleration of the chair is .

Options

A. `0.45 ms^(-2)`
B. zero
C. `(9)/(25 ms^(-2)`
D. `2 ms^(-2)`

Answer

D

Explanation

Let a be the acceleration of the man and the chair and T be the tension in the rope .
for upward motion of the man ,
`T + 450-1000 =((1000)/(10)) a or T= 550 +100 a`
for upward motion of the chair ,
`T- 450 -250 =((250)/(10)) a or T= 25 a + 700`
equating (i) and (ii ) , we get
`550 +100 a = 25 a + 700`
` or 75 a = 700 - 550 = 150 or a=(150 )/( 75 ) = 2 ms^(-2)`