A 0.840-fcg glider on a level air track is joined by stringt to two hanging mass
ID: 1357692 • Letter: A
Question
A 0.840-fcg glider on a level air track is joined by stringt to two hanging masses. As seen in the figure, the mass on the left is 4.85kg and the one on the right is 3.62kg The strings have negligible mass and pass over light, frictionless pulleys. Find the acceleration of the masses when the air flow is turned off and the coefficient of friction between the glider and the track is 0.45. Take positive to be an acceleration to the right. Tries 0/12 Find the tension in the string on the left between the glider and the 4.85-kg mass when the air How is turned off and the coefficient of friction between the glider and the track is 0.45. Find the tension in the string on the right between the glider and the 3.62-kg mass when the air flow is turned off and the coefficient of friction between the glider and the track is 0.45.Explanation / Answer
1) on 4.85 kg net force is 4.85-T1=4.85a----------(1)
on 0.840kg frictional force is acting left side T1-f=0.840a----------(2) here f=mue x N=mue x mg
adding equations 1 and 2
4.85-.45 x0.840 x 10=4.85a+0.840a
a=(4.85-3.78)/5.69
a=0.18m/s^2
2) 4.85-T1=4.85a
T1=4.85-0.873=3.9N
3)T2-3.62 x10=3.62a
T2=36.2+0.65=36.85N
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