A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of `2.5 m//s^(2)` as shown in the figure. The force of interaction between the top link and the link immediately below it, will be

Options

A. 6.15 N
B. 4.92 N
C. 3.69 N
D. 2046 N

Answer

B

Explanation

`because" "F=5mg = 5ma`
F= 5mg + 5ma
`= 5m(g+a)`
If the force of interaction between top (first) link and second is T. Then,
ma = F - mg - T
T = F - mg - ma
`=5 mg + 5ma - mg - ma`
`=4mg + 4ma=4m (g+a)`
`T=4xx0.1 (9.8 + 25) = 4.92 N`