# Previous answers premise the diagram below is a

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5.15/15 points | Previous AnswersPremiseThe diagram below is a schematic of the experimental setup. Note: Motion to the right (+x) and motion in the downward ydirection aretaken as positive in this lab.Part (a)What is the applied force that sets the two masses in motion? Part (b)What is the expression for the net force Fnet1acting on mass m1? (Use the following as necessary: g, m1, Fnet1,x=\$\$0T.)
a
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11/3/2017Lab 3 - Newton's Second Law: PreLab9/15Part (c)What is the expression for the net force Fnet2acting on mass m2? (Use the following as necessary: f, g, m2, Tand Fnet2,x=\$\$TƒFnet2,y=\$\$m2gFNFN.)
Consider the vertical motion of mass m1. In part (b), you wrote down the net force acting on the mass. This is the left hand side of theequation for Newton's Second Law (Fnet= ma) and this net force equals m1a. Similarly, for the horizontal motion of mass m2the left handside of the equation for Newton's Second Law is the net horizontal force and the right hand side is m2aBased on the above explanation and your answers in parts (b) and (c), what is the expression for the acceleration aof the two masses?Eliminate Tand give your answers in terms of m1, m2, g, and f. only.
You have completed the problem!Additional MaterialsNewton's Second LawAppendix
11/3/2017Lab 3 - Newton's Second Law: PreLab10/156.5/5 points | Previous AnswersIn this lab you will fit your data to a straight line. Which variables will be kept constant? (Select all that apply.)Additional MaterialsNewton's Second LawAppendix
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11/3/2017Lab 3 - Newton's Second Law: PreLab11/15