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Use Ko+Uo+W NCF=Kf+U f Answer the following questions about the system as the mass descends and the glider moves. Use only symbols (M1, mh, g, H1, vo, vf) in your response and list: a) The glider’s initial kinetic energy relative to the tabletop. b) The glider’s final kinetic energy relative to the tabletop. c) The mass hanger’s initial gravitational potential energy relative to the floor. d) The mass hanger’s final gravitational potential energy relative to the floor. e) The mass hanger’s initial kinetic energy relative to the floor. f) The mass hanger’s final kinetic energy relative to the floor.

Respuesta :

Answer:

a) K₀ = ½ m v₀² , b) [tex]K_{f}[/tex] = ½ m [tex]v_{f}[/tex]² , c)  U = M g H1 , d) U = 0 , e) K = ½ M v₀² ,

f)   K = ½ M [tex]v_{f}[/tex]²

Explanation:

a) the mass of the glider is m

kinetic energy is

         K = ½ m v²

It's like the table and the floor are fixed, the relative speed  does not changes

         K₀ = ½ m v₀²

b) [tex]K_{f}[/tex] = ½ m [tex]v_{f}[/tex]²

c) the hanging mass is M

   In this case the mass that is hanging is on the table that is at a height H1 with respect to the floor, the initial potential energy is

        U = M g H1

d) the hanging mass decreases in height until it reaches the floor, where its height is zero

        U = 0

e) If we assume that the glider and the hanging mass are connected by a rope, it must be in tension at all times, so the speed of the glider and the mass are the same, since they carry the same acceleration

        K = ½ M v₀²

F)

        K = ½ M [tex]v_{f}[/tex]²

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