As shown above, mass m=6.6 kg slides on a frictionless incline and is attached b
ID: 1776760 • Letter: A
Question
As shown above, mass m=6.6 kg slides on a frictionless incline and is attached by a string to another mass 2m which hangs over a pulley. The incline makes an angle theta=21 degrees with the horizontal. The string connecting to the masses as well as the pulley have negligible mass and friction. What is the tension (in Newtons) in the connecting string? As shown above, mass m=6.6 kg slides on a frictionless incline and is attached by a string to another mass 2m which hangs over a pulley. The incline makes an angle theta=21 degrees with the horizontal. The string connecting to the masses as well as the pulley have negligible mass and friction. What is the tension (in Newtons) in the connecting string?Explanation / Answer
Apply Newon second law to the blocks
T- m2g = m2a
m1gsintheta- T= m1a
m1g sin theta -m2g-m2a= m1a
(m1 sin theta- m2) g/ m1+ m2 = a
a = ( 6.6 sin21 - 2) 9.8/ 6.6 + 2
=0.41 m/s^2
tension in the string is
T = m2( g+ a)
= 2 ( 9.8+ 0.41)
=20.43 N
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