JAMB - Physics (2003 - No. 42)
A bead traveling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1 determine the work done by the friction
\( 2 \times 10^{-4}J \)
\( 2 \times 10^{-3}J \)
\( 2 \times 10^{1}J \)
\( 2 \times 10^{2}J \)
Selitys
Work done by friction = Friction force x Displacement from the relation, f = uR
where u = coefficient of friction; R= Normal reaction
F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)
where u = coefficient of friction; R= Normal reaction
F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)