# Find speed using the work-energy theorem

1. Mar 27, 2015

### Calpalned

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Work energy theorem $\Delta W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$

3. 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?

2. Mar 27, 2015

### 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?

3. Mar 27, 2015

### mattt

$W = W_1 + W_2$

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:

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