Question

A conveyor belt is moving horizontally at a speed of 4 m/s. a box of mass 20 kg is gently laid on it. it takes 0.1 s for the box to come to rest on belt. if the belt continues to move uniformly, then the distance moved by the box on the conveyor belt befor seffling is :

Answer 1

The initial velocity of box relative to the belt is 3m/s in backward direction. (opposite to motion of belt) ie 3 m/s. The box experience maximum frictional force F=μR=μmgF=μR=μmg
Therefore a=Forcemassa=Forcemass
=μmgm=μmgm
=μg=μg
=0.5×9.7=4.9m/s2=0.5×9.7=4.9m/s2
in forward direction
Let the box come to rest in distance S
v2=a2+2asv2=a2+2as
0=(−3)2+2×4.9×s0=(−3)2+2×4.9×s
−9=2×4.9×S−9=2×4.9×S
S=−0.918mS=−0.918m
We sign shows the box moves in opposite direction of motion of belt

Answer 2

Concept:

• One dimensional motion
• Relative motion
• Relative speeds

Given:

• Speed of conveyor belt u = 4m/s
• time for the box to come to rest t = 0.1s
• Mass of box = 20kg

Find:

• Distance moved by the box on the conveyor belt before settling

Solution:

v = u +at

where v is the final velocity which is zero, v = 0

0 = 4 + 0.1a

a = -40m/s^2

s = ut +1/2at^2

s = 4(0.1) + 1/2 (-40)(0.1)(0.1)

s = 0.4 - 0.2

s = 0.2m

The box moves a distance of 0.2m on the conveyor belt.

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