Abox is sliding with a speed of 4.50 m/s on a horizontal surface when, at point p, it encounters a rough section. on the rough section, the coefficient of friction is not constant but starts at .100 at p and increases linerly with distance past p, reaching a value of .600 at 12.5 m past point p. (a) use the work energy theorem to find how far this box slides before stopping. (b) what is the coefficient of friction at the stopping point? (c) how far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?

a) 5.11 m

b) 0.3

c) 10.33 m

Explanation:

a) Since the friction coefficient is linearly proportional to distance, its formula is given by:

Recall that and  :

Replacing A and B in (1):

According to Newton's second law:

The work done by the friction force is:

Work-energy principle states that work is the difference of kinetic energy:

Recall that the final speed is zero. Replacing (3) in (4):

Solving x:

We take the positive root, so the box slides 5.11 m before stopping.

b) We replace 5.11 m in (2):

c) Using (3) and (4) with constant friction coefficient:

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered