Question

A 15kg box is given an initial push so that it slides across the floor & comes to a stop. I the coefficient of friction is .30,
a) find Ffr- Fg=mg - Fg=15kg(9.8N/kg)=147N
Ffr=mu(Fn) - Ffr=.30(147N)=44.1

b) Find the acceleration of the box. Hint: what is the net force as the box slides to a stop?

I really want to say 0 but then c) asks if the initial speed is 3.0 m/s what is d & I'm having a hard time getting anything but 0 w/ the info I have & a=0

Answer 1

In the vertical direction mg down from gravity = mg up from the floor so the vertical acceleration = 0

In the horizontal direction there is only one force, friction, deaccerating
-0.30 m g = m a
so
a = -0.30 g = -0.30 * 9.8 =-2.94 m/s^3

Now if the initial speed is 3
then
v = 3 - a t
v = 3 - 2.94 t
at stop
v = 0
so
0 = 3 - 2.94 t
t = 1.02 seconds to stop
now you can use the distance equation to get distance d
d = 3 t - (1/2)(2.94)t
but it is easier to use average speed = 3/2 = 1.5 m/s for 1.02 sec
1.5 * 1.02 = 1.53 meters :)
v = 3 -

and a =