A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance of 8 m across a level floor. What is the coefficient of kinetic friction between the object and the floor?

Answer 1

average speed = (3+0)/2 = 1.5 m/s

so t = 8/1.5 = 80/15 = 5.33 seconds sliding to a stop

then acceleration a = (0-3)/5.33 = -.5625 m/s^2

so now the physics
F = m a
-m g (mu) = m (-.5625)
mu = .5625/9.81 = .0573

Answer 2

M*g = 0.5 * 9.8 = 4.9 N. = Wt. of object = Normal force(Fn).

V^2 = Vo^2 + 2a*d.
0 = 9 + 2a*8,
a = -0.563 m/s^2.

Fk = u*Fn = u*4.9 = 4.9u = Force of kinetic friction.

Fap-Fk = M*a.
0 - 4.9u = 0.5*(-0.563),
u = 0.057.

Answer 3

from energy=
1/2mv^2.
=1/2×0.5×3^2

=2.25 but

work done=force applied
then force=2.25
recall f=ma
=0.5×10
5N
so.
coefficient=kinetic frictions limiting friction
2.25÷5=0.45
coefficient of kinetic friction =0.45

Answer 4

acceleration is not 10 since it is no a free fall, i think alfred can find alternative solving to attemp in order to have more convenient answer