Physics

I'd guess there is a force acting against the motion, and if c=0.73, it's
F = c·m·g
where g is the gravitational acceleration, and so you got an acceleration
a = F/m = c·g
So
v(t) = v(0) - c·g·t
v(t) = 0 ==> t* = v(0)/(c·g) ==> v(t) = c·g·(t*-t)
x* = int_0^{t*} v(t) dt = c·g·(t*)^2/2 = v(0)^2/(2·c·g)

0 - 1/2*m*v^2 = - .73*m*g*d

>>8619117
1/2*v^2 = (0.73)*g*d

Op here.

Thanks all. I will work out and see if the answer matches the key and I'll post an update.

As promised. Here is my work and the correct answer.

KE (kinetic energy) = 1/2 (1.7kg)(5.1^2) = 22.1 Joules
f = (.73)(1.7kg)(9.8) = 12.2N

22.1J/12.2N = 1.8 meters This is the correct answer

>>8619072
The book never comes to a rest.

>>8619157
It assumes there is a kinetic friction coefficient of .73, so it will eventually have to come to rest because friction is going to forever act on the book.

>>8619157
I think they are assuming a hypothetical frame of reference where things can come to rest

It's not possible to solve.
I take this question to mean that the kid pushed an already moving book. We have no way of calculating the force he put on it.

>>8619324
I think it's just worded wrong, my old physics textbook was littered with shit like this.