Question

A 32 kilogram box slides along a level surface. If μk between the box and the surface is 0.68, determine the acceleration of the box. Assume "forward" is positive.

1. what is the acceleration?____ m/s²


Suppose the box had an initial velocity of +1.310 m/s. How far will the box slide before it stops?

2.____m

Answer 1

1:Acelleration is equal to 6.664m/s^2

2:-0.128m

Step-by-step explanation:

1:Use the fact that Ff=mew K * FN

=32*9.8=Fn

=0.68=mew k

=213.248=6.664m/s^2

2:Use your eqn Displacement=vf^2-vi^2/2a

Given that vf=0

-(1.310)/2(6.664)

=-0.128m