Question
Question: A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration ...
A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of 2.5 m/s2. The force of interaction between the top link and the link immediately below it, will be –
A
6.15 N
B
4.92 N
C
3.69 N
D
2046 N
Answer
4.92 N
Explanation
Solution
T = mg + ma
= 0.4 g + 0.4 a
= 0.4(g + a)
= 0.4(9.8 + 2.5) = 4.92 N