Question

A uniform chain has a mass m and length l it is held on a frictionless table with one sixth of its legth hanging the work done in just pulling the hanging part back on the table is

Answer 1

Answer:

20

Explanation:

U=-25

M1=f/f-u

M1=f/f-(-25)=f/f+25

U2=-40cm

M2=f/f-u2=f/f-(-40)

M2=f/f+40)

M1/M2=4

F+40/F+25=4

=4f+100=f+40

=3f=-60

F=20