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?

2 Answers
Aug 6, 2015

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

Explanation:

Have a look:
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Aug 6, 2015

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")#