First slide

Kinetic friction

Question

A heavy block of mass M is slowly placed on a conveyer belt moving with a speed v. The coefficient of friction between the block and the belt is µ. Through what distance will the block slide on the belt?

Moderate

A

v/µg
B

v²/µg
C

v/2µg
D

v²/2µg

Solution

Force of friction = µ M g
$∴$ Acceleration = µ g
Now $v^{2} = u^{2} + 2 as$
The block will slide on the belt till its velocity becomes v.
Hence
$v^{2} = 2 μgs or s = (v^{2} / 2 μg) .$

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