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 m1 falls down 2.6 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.)
U1 =

(b) 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.)
U2 =

  

(c) Use m1 = 5 kg, m2 = 7 kg, and = 28°. Calculate the total change in potential energy Usystem of the system. Note: A positive answer indicates an increase in potential energy and a negative answer indicates a decrease in potential energy.
Usystem =  

(d) What is the total change in kinetic energy Ksystem of the system. Note: A positive answer indicates an increase in kinetic energy and a negative answer indicates a decrease in kinetic energy.
Ksystem =  

(e) What is the final velocity v of the system?
v =  

(f) 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.)
Conservation of Energy involves only scalar quantities.Both methods have the same level of complexity.Conservation of Energy method does not require free body diagrams showing the force vectors.We have to find the components of the force vectors regardless of the method used.Using Conservation of Energy is simpler.

Explanation / Answer

here,

(a)

If m1 falls down 2.6 m

the change in potential energy delta U1 of m1 , delta U1 = - m1 * g * 2.6

the change in potential energy delta U1 of m1 is - 2.6 m1g

(b)

If m1 falls down 2.6 m

m2 raises , h = 2.6 * sin(theta)

the change in potential energy delta U1 of m1 , delta U1 = m2 * g * 2.68sin(theta)

the change in potential energy delta U1 of m1 is 2.6 m2g * sin(theta)

(c)

m1 = 5 kg

m2 = 7 kg

theta = 28 degree

the change in potential energy of the system , PE = 2.6 m2g * sin(theta) - 2.6 m1g

PE = 2.6 * 7 * 9.8 * sin(28) - 2.6 * 5 * 9.8

PE = -43.67 J

the change in potential energy of the system is - 43.67 J

(b)

the total energy of the system remains conserved

kinetic energy = change in potential energy

the total change in kinetic energy of the system is 43.67 J

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