A bead travelling on a straight line wire is brought to rest at 0.2 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
[g = 10ms\(^2\)]
a
2 x 10\(^{-4}\)
b
2 x 10\(^{-3}\)
c
2 x 10\(^{-1}\)
d
2 x 10\(^2\)
Explanation
Correct Option
bVideo Explanation
No video available
Post your Contribution
Share:
Discussions (8)
olyezema
4 years ago
the answers is very correct
μ=F/R
F=μ×R
R=0.01×10
R=0.1N
F=0.1×0.1
F=0.01
w=0.01×0.2
w=0.002j
w=2×10^-3j
OG_Loc
2 years ago
The Work done by friction = Frictional force x displacement
f = μR
μ = coefficient of friction ; R = normal reaction
=> F = μ mg (R = mg)
= 0.1 x 0.01 x 10
W = F x displacement
= 0.1 x 0.01 x 10 x 0.2
= 2 x 10−3 J
