# A 20 N crate starting at rest slides down a rough 5m long raml ,inclined at 25° with the horizontal. 20 j of energt is lost to friction. What will be the speed of the crate at the bottom of the incline?

Aug 6, 2015

I found: $4.4 \frac{m}{s}$ (but check carefully math and method).

#### Explanation:

Have a look:

Aug 6, 2015

Gio already wrote something, but just to have an answer in MathJAX (:P), I'll write my approach too.

$E = \Delta K + \Delta U - {E}_{{F}_{s}} = 0 = \frac{1}{2} m {v}_{f}^{2} - m g {y}_{i} - 20 \text{J}$
...while ${y}_{f} = 0$ and ${v}_{i} = 0$. We want ${v}_{f}$.

($K$ is kinetic energy, $U$ is potential energy, and ${E}_{{F}_{s}}$ is energy gained/lost due to static friction since the crate is sliding.)

All constants are positive in this context, so a negative either means down is negative or losing energy is negative. Since this ramp is $5$ meters "long" along the surface, and we need ${y}_{i}$:

$\sin \left({25}^{o}\right) = \frac{{y}_{i}}{5} \implies {y}_{i} = 5 \sin \left({25}^{o}\right)$

The weight is just the force due to gravity: ${F}_{g} = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{- m g = 20 \text{N") => color(green)(-20/g "kg} = m}$

So overall:

$20 \text{J} = \frac{1}{2} \textcolor{g r e e n}{m} {v}_{f}^{2} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{- m g} {y}_{i}$

$= \frac{1}{2} \cdot \textcolor{g r e e n}{- \frac{20}{g}} {v}_{f}^{2} + \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{20} \cdot 5 \sin \left({25}^{o}\right)$

$= - \frac{10}{9.08665} {v}_{f}^{2} + 42.26 \text{J}$

$22.26 \text{J} = \frac{10}{9.08665} {v}_{f}^{2}$

$21.8296 {\left(\text{m/s}\right)}^{2} = {v}_{f}^{2}$

$\textcolor{b l u e}{{v}_{f} \approx 4.672 \text{m/s}}$