Question

Find average friction?

As shown on the figure, a bead of mass 15 g slides on an wire.  The speed of the bead at point A is 6 m/s.  After travelling a distance of 250 cm from point A to point the C, the bead stops at point C.  What is the average friction experience by the bead in travelling from point A to point C.

Answer 1

Average Frictional Force = 0.126 N

Explanation:

Frictional losses cause an energy loss = Loss of KE + Loss of PE
Work done by friction = F(fr) distance = m a d
Loss of KE = 1/2 m v^2 = 1/2 .015 6^2 = 0.27 Joules
Loss of PE = m g (change in height) = .015 9.8 0.3 = .0441 Joules
d = 250 cm = 2.5 m
Frictional loss of energy = F(fr) d
F(fr) d = 0.27 + 0.0441 = 0.3141 Joules
F(fr) = .3141 / 2.5 = 0.126 Newtons of Friction

Answer 2

Energy of bead at point `A`
PE`+`KE `=E_A=mgh_A+1/2mv^2`

Energy of bead at point `C`
`E_C=mgh_C`

Loss in energy while bead moves from `A` to `C`
`E_A-E_C=mgh_A+1/2mv^2-mgh_C`
`=>E_A-E_C=mg(h_A-h_C)+1/2mv^2`

Taking `g=9.81ms^-2` and inserting given values we get

`=>E_A-E_C=0.015xx9.81xx0.30+1/2xx0.015xx6^2`
`=>E_A-E_C=0.0441+0.27=0.3141J`

Using Law of conservation of energy, this loss is equal to work done against force of friction.

Work done against force of friction `= "Force"_fxx "distance"`
If `F` is average force of friction work done in moving through a distance of `250cm`

`Fxx2.5= 0.3141`
`F=0.3141/2.5= 0.13 N`, rounded to two decimal places.