Use the exact values you enter to make later calculations. The set up in the dia

Question

Use the exact values you enter to make later calculations. The set up in the diagram below relates to a classic inclined plane problem that is typically solved using free body diagrams and Newton's Second Law of Motion. You will work through this inclined plane problem using Conservation of Mechanical Energy instead. You may ignore friction and assume that the system is initially at rest (a) If mi falls down 2.9 m, what is the change in potential energy AU1 of m1? Leave your answer in terms of mi and g. Note: A positive answer indicates an increase in potential energy and a negative answer indicates a decrease in potential energy. (Assume any numerical value for height is in meters, but do not enter units.) 2.9m AU Notice that the string connecting mi and does not m2 stretch but remains taut. What is the responding change in potential energy AU2 of m2 (b) Leave your answer in terms of m2, g, and 6. Note: A positive answer indicates an increase in potential energy and a negative answer indicates a decrease in potential energy (Assume any numerical value for height is in meters, but do not enter units. mog2.9sin (0) AU kg, and 6 Calculate the total change in potential energy AU (c) Use m kg, m2 f the system Note: A positive answer indicates an increase in 30 system potential energy and a negative answer indicates a decrease in potential energy. AU 131.32 J system (d) What is the total change in kinetic energy AKS of the system. Note: A positive answer indicates an increase in kinetic energy and a negative answer indicates a system decrease in kinetic energy. syste 397.96 J (e) What is the final velocity v of the system? 7.54 m/s

Explanation / Answer

C)
change in potential energy of the system is deltaU_sysytem = deltaU1+deltaU2

deltaU_system = (-2.9*m1*g)+(m2*g*2.9*sin(theta))

deltaU_system = (-2.9*6*9.81)+(8*9.81*2.9*sin(30)) = -56.898 J

D) total change in kinetic energy = -deltaU_system = 56.898 J

e) total change in kinetic energy = 0.5*(m1+m2)*vf^2 = 56.898

final velocity of the system is vf = sqrt(56.898/(0.5*(m1+m2)))

vf = sqrt(56.898/(0.5*(6+8))) = 2.85 m/s

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