A box slides down a plank, Find its velocity

12 replies · 2K views · last reply
Girn261
Attachments
  • Handwritten physics working on lined paper: an inclined plank triangle labelled 1.2 m and 0.6 m with angle 26.6 degrees, then a = g(sin θ - μ cos θ) = 0.8829 m/s², hypotenuse T = 1.346 m, and v² = u² + 2as giving v = 1.54.
    20170819_135458_HDR.jpg
    14.5 KB · Views: 408
haruspex said:
1.2 m length of plank.

hmm, still get wrong answer.. With 1.2m being the length of the incline, I get 1.039 for the "x" length. and with V being 1.3544
 
Last edited by a moderator:
Please show your working. I solved it two different ways (one of which was your method using plank length = 1.2) and both give same answer (which wasn't 1.3544). So most likely an arithmetic error somewhere.
 
CWatters said:
Please show your working. I solved it two different ways (one of which was your method using plank length = 1.2) and both give same answer (which wasn't 1.3544). So most likely an arithmetic error somewhere.
1.2^2=a^2+0.6m^2
a=1.039

v^2=u^2+2(a)(s)
0+(2)(.8829)(1.039)
sqrt 1.834 = 1.3544
 
Girn261 said:
1.2^2=a^2+0.6m^2
a=1.039
v^2=u^2+2(a)(s)
0+(2)(.8829)(1.039)
sqrt 1.834 = 1.3544

It's confusing using "a" for two things.

The acceleration isn't 0.8829 m/s/s. Show how you got that.

The distance the object moves isn't 1.039m it's 1.2m.

Sorry got to go to bed now. It's nearly midnight here.
 
ATTACH]
Handwritten physics working on lined paper: a sketch of a box on an incline with a 1.2 m base and 0.6 m height, tan θ = 0.6/1.2, then a = g(sin θ − cos θ μ) = 0.8829 m/s², T² = 1.2² + 0.6², T = 1.346, and v² = u² + 2as giving 1.54.
CWatters said:
It's confusing using "a" for two things.

The acceleration isn't 0.8829 m/s/s. Show how you got that.

The distance the object moves isn't 1.039m it's 1.2m.

Sorry got to go to bed now. It's nearly midnight here.

Please see picture for how I got acceleration
 
Girn261 said:
ATTACH]
Handwritten working on lined notebook paper: a small ramp diagram labelled 0.6 m high and 1.2 m long with tan θ = 0.6/1.2 = 26.57°, then a = g(sin θ - cos θ μk) = 0.8829 m/s², T² = 1.2² + 0.6² giving T = 1.346, and v² = 0 + 2(0.8829)(1.346) giving 1.54.


Please see picture for how I got acceleration
This seems to be your first picture again, with the wrong plank length.
 
haruspex said:
This seems to be your first picture again, with the wrong plank length.

I know, just quoting to show how I got acceleration. With plank as A or B in pythagorean theorem, I still get wrong answer
 
Girn261 said:
I know, just quoting to show how I got acceleration. With plank as A or B in pythagorean theorem, I still get wrong answer
Please post your working for how you get the wrong acceleration eventhough using the right plank length.
And, please, not as an image. Take the trouble to type your algebra so that people can read it much more easily and quote parts to point out errors.
 
  • Like
Likes   Reactions: CWatters
Girn261 said:
I know, just quoting to show how I got acceleration.

The equation for the acceleration contains terms Sin and Cos, you need to redo these calculations because they are also affected by the error with the plank length.
 
I figured out my mistakes. Thank you for everyones help, I really appreciate this forum.
 
What answer did you get?

Just for info, the alternative method I used was to apply conservation of energy.

The final KE = Initial PE - Wf

where
Wf = the work done against friction = L * μmgCosθ
L = length of plank

so..

0.5mv2 = mgh - μmgLCosθ

mass cancels.
Rearrange to give

v = SQRT(2g(h - uLCosθ))