+1 vote
in Machine Kinematics by (38.0k points)
A heavy block of mass m is slowly placed on a conveyer belt moving with speed v. If coefficient of friction between block and the belt is μ, the block will slide on the belt through distance

(a) v/μg

(b) v^2/√μg

(c) (v/μg)^2

(d) v^2/2μg

The question was posed to me in semester exam.

Query is from Coefficient of Friction topic in chapter Friction of Machine Kinematics

1 Answer

0 votes
by (106k points)
Right option is (d) v^2/2μg

For explanation: Retardation due to friction force = μg

V^2 = 2.μg s

s = v^2/2μg.

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