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.
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. If m1 falls down 2.8 m, what is the change in potential energy ?U1 of m1? Leave your answer in terms of m1 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.) Notice that the string connecting m1 and m2 does not stretch but remains taut. What is the corresponding change in potential energy ?U2 of m2? Leave your answer in terms of m2, g, and ?. 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.) Use m1 = 5 kg, m2 = 6 kg, and ? = 26 degree. Calculate the total change in potential energy ?U system of the system. Note: A positive answer indicates an increase in potential energy and a negative answer indicates a decrease in potential energy. What is the total change in kinetic energy ?K system of the system. Note: A positive answer indicates an increase in kinetic energy and a negative answer indicates a decrease in kinetic energy. What is the final velocity v of the system? What are the advantages in using Conservation of Energy to solve this problem instead of Newton's Second Law of Motion? (Select all that apply.)
Explanation / Answer
a) delta U1 = -m1*g*h
b) delta U2 = +m2*g*h*sin(theta)
c)
delta U1 = -5*9.8*2.8 = -137.2 J
delta U2 = 6*9.8*2.8*sin(26) = +72.17 J
delta U(system) = delta U1 + delta U2 = 137.2 + 72.17 = -65.03 J
d)
delta U = - delta K = +65.03 J
e) 0.5*(m1+m2)*v^2 = 65.03
v^2 = 65.03*2/(m1+m2)
v = sqrt( 65.03*2/(5+6))
v = 3.45 m./s
