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An object of mass 12.9 kg enters a rough floor with speed of 10.1 m/s. The coefficient of friction between the floor and the object is 0.30 and the object moves 2.07 m before returning to a smooth surface (frictionless). What is the velocity of the object once it returns to the smooth surface

Answer :

hamzaahmeds

Answer:

vf = 9.48 m/s

Explanation:

From the law of conservation of energy we can write:

[tex]Kinetic\ Energy\ Lost = Work\ Against\ Friction\\\frac{1}{2}m(v_i^2 - v_f^2) = fd\\\frac{1}{2}m(v_i^2 - v_f^2) = (\mu W)d\\\frac{1}{2}m(v_i^2 - v_f^2) = (\mu mg)d\\\frac{1}{2}(v_i^2 - v_f^2) = \mu gd\\v_f^2 = v_i^2 - 2\mu gd[/tex]

where,

vf = final speed = ?

vi = initial speed = 10.1 m/s

μ = coefficient of friction = 0.3

g = acceleration due to gravity = 9.81 m/s²

d = distance covered = 2.07 m

Therefore,

[tex]v_f^2 = (10.1\ m/s)^2 - 2(0.3)(9.81\ m/s^2)(2.07\ m)\\v_f^2 = 102.01\ m^2/s^2 - 12.18\ m^2/s^2\\v_f = \sqrt{89.83\ m^2/s^2}\\[/tex]

vf = 9.48 m/s

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