Question

A 15 kg box is given an initial push so that it slides across the floor and comes to a stop. If the coefficient of friction is 0.30, a) find the friction force.
b) find the acceleration of the box. Hint: what is the net force as the box slides to a stop?
c) how far will the box go if its initial speed is 3.0 m/s?

(I only need help on part c)

Answer 1

Answer:

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