Question

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?

Answer 1

I found: `4.4m/s` (but check carefully math and method).

Explanation:

Have a look:

Answer 2

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

I would start with:
`E = DeltaK + DeltaU - E_(F_s) = 0 = 1/2mv_f^2 - mgy_i - 20 "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(25^o) = (y_i)/5 => y_i = 5sin(25^o)`

The weight is just the force due to gravity: `F_g = color(darkorange)(-mg = 20 "N") => color(green)(-20/g "kg" = m)`

So overall:

`20 "J" = 1/2color(green)(m)v_f^2 color(darkorange)(- mg)y_i`

`= 1/2*color(green)(-20/g)v_f^2 + color(darkorange)(20 )*5sin(25^o)`

`= -10/(9.08665)v_f^2 + 42.26 "J"`

`22.26 "J" = 10/(9.08665)v_f^2`

`21.8296 ("m/s")^2 = v_f^2`

`color(blue)(v_f ~~ 4.672 "m/s")`