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A block of mass m = 7.20 kg is released from rest from point circled A and slides on the frictionless track shown in the figure below. (Assume ha = 7.70 m.) (a) Determine the block's speed at points circled B and circled C. point circled B m/s point circled C m/s (b) Determine the net work done by the gravitational force on the block as it moves from point circled A to point circled C. J
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Added Thu, 06 Aug '15
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TEa = TEb

m*g*ha +0.5*m*Va^2 = m*g*hb+ 0.5*m*Vb^2

m*g*(ha-hb) = 0.5*m*(Vb2 - Va^2)

Vb = sqrt(2*g*(ha-hb))=sqrt(2*9.8*(7.7-3.2))= 9.39m/s

TEb = TEc

m*g*hb +0.5*m*Vb^2 = m*g*hc+ 0.5*m*Vc^2

m*g*(hb-hc) = 0.5*m*(Vc^2 - Vb^2)

Vc^-9.39^2 = 2*9.8*(3.2-2)

Vc = 10.57 m/s

b) work done = change in KE or change in PE

work = m*g*(ha-hc) = 0.5*m*(Vc^2-Va^2)

W = 7.2*9.8*(7.7-2) = 0.5*7.2*(10.57^2-0^2)

W = 402.2 = 402.2 J
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Added Thu, 06 Aug '15
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