- #1
questtosuccess
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Homework Statement
A 15 kg block slides along a horizontal frictionless surface 3.0 m above the ground at a constant speed of 2.0 m/s. The block then slides down an incline that makes an angle of 35° with the horizontal and has a coefficient of kinetic friction equal to 0.30. After reaching the end of the incline, the box continues sliding horizontally across the frictionless ground. Calculate the kinetic energy of the crate as it slides
a. along the upper surface.
b. along the ground.
Homework Equations
for a) KE= 1/2 mv2
for b) E1 = E2 + Wnon-conservative
The Attempt at a Solution
for a) KE=1/2 mv2 = 1/2 (15) (2)2 = 30 J
for b) I'm not quite sure where to consider the box sliding the ramp is at the moment... at the top (where y = 3 m) or the bottom (where y = 0 m)... regardless, this is how I did it.. but I'm pretty sure it's wrong
I began by finding the speed of the box when it's traveling down the ramp. I compared the energy of the box traveling along the upper surface, and the energy of the box going down the ramp (considering it to be at y = 0 m).
E1 = E2 + Wnon-conservative
1/2mv12 + mgy1 = 1/2mv22 + Ffdcos180
v2 = [itex]\sqrt{}[/itex] (2(mv12/2+ mgy1 - [itex]\mu[/itex]k FNd (-1) )) /m
From here, I solved and found that v2 = 8.79 m/s
I had a problem already though.. I wasn't sure what "d" was supposed to be in this case ... for [itex]\mu[/itex]k FNd cos180 ... I let d = 3m (is that right?)
Finally, I used the above info and compared the energy of the box going down the ramp (*** considering it to be at y = 3 m this time), and the energy of the box traveling along the lower horizontal surface.
E2 + Wnon-conservative = E3
mgy2 + 1/2mv22 + Ffdcos180 = 1/2mv32
Therefore,
KE3 = mgy2 + 1/2mv22 + Ffdcos180
I solved and got KE3 = 910 J
All of this seems very incorrect. I feel like I'm doing this all wrong. Please help ! Thank you.