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## Homework Statement

A 96-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.25. What is the final speed of the crate?

## Homework Equations

Work energy theorem ##\Delta W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ##

## The Attempt at a Solution

- First I'll calculate the speed at the end of the first fifteen meters (when there is no acceleration).

##F \cdot d = (350)(15) = 5250 = ## work

##= \Delta E_k = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2 ##

Because the initial velocity is zero, we see that ## = \frac{1}{2}mv_f^2 ##

##10500 = mv^2 ## so ##v_f = 10.458 ## m/s.

- Now I'll calculate the answer (speed at 30 meters of displacement).

Due to the presence of friction, kinetic and potential energies are not the only energies present.

##E_K + E_P + E_f = ## where the subscript f refers to friction

##\frac{1}{2}mv_0^2+mgh+F_N\mu kl = \frac{1}{2}mv_f^2+mgh+F_N\mu kl ##

##v_0 = 10.458##

##5250+0+0 = \frac {1}{2}mv_f^2 + 0 + 3528 ##

##v_f = 5.98## m/s

The correct answer is twice what I got. What did I do wrong?