Question

A block is gently placed on a long conveyor belt
moving with 11 m s-1. If the coefficient of friction
between block and belt is 0.4, then the block will
slide on the belt up to distance of
a) 1021 m
b) 15.43 m
c) 20.3 m
d) 25.6 m​

Answer 1

Answer:

b)15.4 3 m

Explanation:

Given:

Velocity of conveyor belt, u=11 m/s

coefficient of friction between the block and the belt ∞∞$\mu =0.4$

Initially the velocity of the block will be same as that of the conveyor belt i.e. u=11 m/s

The frictional  force acing on the block causes deceleration

The deceleration of he block is given by $=-\mu g\\=0.4\times9.8\\=-3.92 m/s^2$

Let S be the distance traveled by the block on the belt before coming to rest

So from the equation of the motion we have

$2as=V^2-u^2\\2\times(-3.92)\times S=0^2-11^2\\S=15.43\ \rm m$

Hence the distance traveled is calculated.