Question

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 1

vf = 9.48 m/s

Step-by-step explanation:

From the law of conservation of energy we can write:

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,

vf = 9.48 m/s