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If F = 40 N and M = 1.5 kg, what is the tension in the string connecting M and 2M? Assume that all surfaces are frictionless.

(Look at figure given/attched image)

a. 13 N

b. 23 N

c. 36 N

d. 15 N

e. 28 N

If you apply Newton's Second Law to this problem the sum of the external forces acting on the system will be equal to the mass of the system multiplied by the rate of the resulting acceleration. In this case there are two relevant external forces; the applied force Fa and the weight of the hanging mass Fg. Therefore, Newron's Second Law becomes:

SF=m*a

Fa-Fg=(2M+M)*a

And the resulting acceleration becomes:

a=(Fa-Fg)/(2M+M)=(40-1.5*9.8)/(3*1.5)=25.3/4.5=5.62m/s^2

Now that you know the acceleration of the system as a whole you can now apply Newon's Second Law to just the hanging mass for which the relevant external forces will be the force of gravity Fg and th force of tension Ft. Using Newton's Second Law:

SF=m*a

Ft-Fg=m*a

solving for the tension in the string between the two masses:

Ft=m*a+Fg=1.5*5.62+1.5*9.8=23.1N

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.