Problem 1: (20 pts) A 4.00kg mass rests on a ramp inclined at 10°. An anchored s
ID: 2035961 • Letter: P
Question
Problem 1: (20 pts) A 4.00kg mass rests on a ramp inclined at 10°. An anchored spring (90N/m) is attached below it while a string is attached to the mass which parallels the ramp up the ramp over a pulley and vertically down to a 3.00kg mass. The spring is initially unstretched when the masses are released from rest. The coefficient of friction between the mass and the ramp is 0.29. After the hanging mass has fallen a distance of 20cm, a,) (4 pts) what is the change in gravitational PE of the system throughout the motion? k,? 2. e- Io 2. b) o pty) What s the change in elastic epring potential energs zm c) (4 pts) What is the work done by friction? ?11 .ZI d.) (6 pts) What is the change in kinetic energy throughout the motion? e.) (3 pts) What is the velocity of the masses at this point? NogExplanation / Answer
initial gravitational potential energy of the system Ugi = 0
final gravitational potential energy of the system Ugf = m1*g*h*sintheta - m2*g*h
change in gravitational potential energy dUg = Uf - Ui = m1*g*h*sintheta - m2*g*h
change in gravitational potential energy dUg = 4*9.8*0.2*sin10 - 3*9.8*0.2 = -4.52 J
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(b)
initial elastic potential energy Uei = 0
final elastic potential energy Uef = (1/2)*k*h^2
change in elastic potential energy dUe = Uef - Uei = (1/2)*k*h^2
change in elastic potential energy dUe = (1/2)*90*0.2^2 = 1.8 J
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(c)
friction force fk = uk*m1*g*costheta
work done by friction force = fk*h*cos180 = -fk*h
work done by friction force = -0.29*4*9.8*cos10*0.2 = -2.24 J
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(d)
from work energy relation
total work done = change in KE
total work = Wg + We + Wf
total work = -dUg - dUe + Wf
total work = 4.52 - 1.8 - 2.24 = 0.48 J
change in KE = 0.48 J
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part(e)
change in KE = (1/2)*m1*v^2 + (1/2)*m2*v^2 = (1/2)*(m1+m2)*v^2
(1/2)*(m1+m2)*v^2 = 0.48
(1/2)*(4+3)*v^2 = 0.48
v = 0.37 m/s