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The fig. shows a 2 kg block attached to 0.5 kg mass by a light in-extensible string which passes over a pulley. The force of...

      

The fig. shows a 2 kg block attached to 0.5 kg mass by a light in-extensible string which passes over a pulley. The force of friction between the horizontal bench and block is 3N. The block is released from rest so that both masses move through a distance of 0.6m. Calculate the velocity of the string.
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Answers


Wilfred
Propelling force= 0.5 x 10= 5N
Opposing force= frictional force = 3N
Net Force = 5-3 = 2N
Using F= ma
2= (2+ 0.5)a
a= 0.8m/s2
s= 0.6m
u=0
V2 = U2 + 2as
= 0+2+0.8+0.6
= 0.96

Therefore V= 0.9798
V= 0.098m/s
Wilfykil answered the question on March 26, 2019 at 11:20


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