- #1
bmx_Freestyle
- 11
- 0
There is a mass at the bottom of an inclined plane. It travels with an initial velocity up the inclined plane at an angle θ. There is a coefficient of friction on the ramp. How far up the ramp will the mass go before stopping? What is the speed of the block when it returns to the bottom of the ramp? What percent of the initial total mechanical energy was lost during the mass's trip (going up and then back down?
m=5 kg
vo=40 m/s
θ=30°
S=the distance you are looking for
Coefficient of friction (μ) = 0.15
Work energy theorem=mg(hf-ho) + 1/2 m (vf^2-vo^2) +fs
Attempt:
i set up the work energy theorem and simplified it down to "work=mghf-1/2mvo^s+μ
mgs" and solved for s
and then i used "work= -mgho + 1/2mvf^s +μ
mgs to solve for vf
i honestly had no clue what to do for the third part of this problem
I don't think my answers are right bc i got 0.598 m fr the first part and 2.23 m/s fr the second part...and i couldn't figure out the third part
Help would be appreciated. Thank u very much to all!
m=5 kg
vo=40 m/s
θ=30°
S=the distance you are looking for
Coefficient of friction (μ) = 0.15
Work energy theorem=mg(hf-ho) + 1/2 m (vf^2-vo^2) +fs
Attempt:
i set up the work energy theorem and simplified it down to "work=mghf-1/2mvo^s+μ
mgs" and solved for s
and then i used "work= -mgho + 1/2mvf^s +μ
mgs to solve for vf
i honestly had no clue what to do for the third part of this problem
I don't think my answers are right bc i got 0.598 m fr the first part and 2.23 m/s fr the second part...and i couldn't figure out the third part
Help would be appreciated. Thank u very much to all!