The set up in the diagram below relates to a classic inclined plane problem that

Question

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 2.0 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 = 6 kg, m2 = 8 kg, and = 26°. 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 =

Explanation / Answer

a) delta U1 = -2.2*m1*g

b) m1 is hanging, m2 is on the inclined plane. m1 falls 2.2m. That means m2 moves up the ramp 2.2m
The HEIGHTof m2 increases 2.2*sin(theta). The potential energy of m2 therefore increases
(m2*g)*2.2sin(theta)

c) m1 falls 2.2m and therefore has -6*9.8*2.2 = -129.36 J

m2's PE increases 8*9.8*2.2sin26 =75.61 J

The total change then is=75.61-129.36=-53.75 J

d) To find m1's velocity after falling 2.2m
Use 53.75=1/2m*v^2

v=2.77 m/sec

Assuming the inclined plane is frictionless
Since m2 is roped to m1, it's velocity up the ramp will be 2.77m/s too

The total increase in KE is then 0.5[8+6]*2.77^2 =53.71 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.