Find speed using the work-energy theorem

[Calpalned]

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?

 

[robphy]

(There is acceleration during the first 15 m... there is no friction, however.)
During the next 15m, is it still being pulled by that same applied force of 350N?

 

[mattt]

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?

$W = W_1 + W_2$

W_1 (first 15 meters) and W_2 (second 15 meters)

$W=\Delta E_K=\frac{1}{2}m v_f^2$ (given that v_0 = 0 )

So:

$\frac{1}{2}mv_f^2=W_1 + W_2 = F.d + (F-R).d$

Being $m=96, F=350, d=15, R=0.25*m*9.8$So you have:

$v_f = \sqrt{\frac{2(2F-R)d}{m}}$