a. Take the system to be mass 1 and mass 2 (thestring is light and the pulley is

Question

a. Take the system to be mass 1 and mass 2 (thestring is lightand the pulley is massless). Ifboth masses are released from rest, what is the change in kineticenergy of the system after mass 2 has descended a heighth. Your answer can only be in term ofm1,m2,v1,v2, and any necessaryconstants where v1 and v2 are the speed of the masses at the instant that mass2 has descended to a point a distance h below whereit starts.

c. In this same time period,what is the change in gravitational potential energy of the system?Your answer can only be in terms of m1, m2, h, and anynecessary constants.

Explanation / Answer

(a) Since m1 doesnot change its vertical position, there isno change in its gravitational potential energy. Hence, the only change in potential energy is due to descendof m2. Gravitational force, displacement, taking upward positiveare : F2 = -m2g and s = -h, => W = -DU = DK = (-m2g)*(-h) = m2gh (b) Since the string is light, the pulley is massless, the speedsv1 and v2 have to be equal to prevent extension of string since itis not mentioned being elastic. (c) Change in gravitational potential energy is : DU = -m2gh asfound earlier in a. (d) Total Kinetic energy = DK - 0 = 0.5*m1*v1*v1 + 0.5*m2*v2*v2 =m2gh => v1=v2=v(suppose) => (m1+m2)*0.5*v*v = m2*g*h => v = [(2m2gh)/(m1+m2)]

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