Mechanics: Friction - 200kg Bag Winched up 10m Slope at 6°

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Discussion Overview

The discussion revolves around a physics problem involving a 200kg bag being winched up a 10m slope at a 6-degree angle, with a coefficient of friction of 0.4. Participants are attempting to determine the distance the bag has moved when its speed decreases from 2 m/s to 1.5 m/s, using principles of mechanics, including forces, acceleration, and energy conservation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the acceleration and distance using kinematic equations and expresses uncertainty about the correctness of their answer compared to the textbook solution.
  • Another participant presents a formula for net work and kinetic energy, agreeing with the textbook solution but indicates they have not yet studied the chapter on kinetic energy.
  • Several participants repeatedly suggest using the same formula for acceleration and distance, indicating a potential method to solve the problem.
  • A participant questions whether the 10m length should be interpreted as height, suggesting a possible misunderstanding of the problem setup.
  • Another participant expresses gratitude after resolving their confusion, indicating a collaborative atmosphere in seeking clarification.

Areas of Agreement / Disagreement

There is no consensus on the correct approach or solution to the problem, as multiple methods and interpretations are presented, and participants express varying levels of understanding and agreement with the textbook answer.

Contextual Notes

Some calculations depend on assumptions about the definitions of distance and height, as well as the application of friction and forces. There are unresolved mathematical steps and differing interpretations of the problem's parameters.

Shah 72
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A bag of sand of mass 200kg is being winched up a slope of length 10m which is at an angle of 6 degree to the horizontal. The slope is rough and the coefficient of friction is 0.4. The winch provides a force of 1000N parallel to the slope. At the bottom of the slope the bag is moving at 2 m/s. Find the distance it has moved when it's speed has reduced to 1.5m/s
R=2000cos6= 1989N
F=1000-(0.4×1989+2000sin6),
By using F=m×a, I get a=-0.023m/s^2
V^2=u^2+2as, u=2m/s, s=10m, I get v=1.88m/s.
Now for u=1.88 and v=1.5 and using the same v^2= u^2+2as, I get s=27.9. So total distance will be 37.9. But the textbook ans is 37.4m. Pls advise if my ans is correct
 
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$W_{net} = \Delta KE = \dfrac{1}{2}m(v_f^2-v_0^2) = F_{net} \cdot \Delta x \implies \Delta x = \dfrac{\Delta KE}{F_{net}}$

$F_{net} = 1000 - mg(\sin{\theta} + \mu \cos{\theta})$

I agree with the text solution ...
 
skeeter said:
$W_{net} = \Delta KE = \dfrac{1}{2}m(v_f^2-v_0^2) = F_{net} \cdot \Delta x \implies \Delta x = \dfrac{\Delta KE}{F_{net}}$

$F_{net} = 1000 - mg(\sin{\theta} + \mu \cos{\theta})$

I agree with the text solution ...
I haven't still done the chapter of KE. So I don't know to apply the formula. If you can pls pls advise using the coefficient of friction and Newtons law.
Thank you so much!
 
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$

try again ...
 
skeeter said:
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$

try again ...
Sure I will try. Thanks a lotttt!
 
skeeter said:
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$

try again ...
m=200kg, s=10m, coefficient of friction =0.4 and initial velocity =2m/s
By using Newtons law
F=m×a
1000-[0.4x2000cos 6+2000sin6) =200a
a=-0.023m/s^2. Iam not getting the ans. If you can pls pls help.
 
skeeter said:
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$

try again ...
I tried and I got the ans. Thank you!
 
A bag of sand of mass 200kg is being winched up a slope of length 10m which is at an angle of 6 degree to the horizontal. The slope is rough and the coefficient of friction is 0.4. The winch provides a force of 1000N parallel to the slope. At the bottom of the slope the bag is moving at 2 m/s. Find the distance it has moved when it's speed has reduced to 1.5m/s

Maybe that 10m "length" should be height? Otherwise, the given solution makes no sense.
 
skeeter said:
Maybe that 10m "length" should be height? Otherwise, the given solution makes no sense.
Thanks I got it. Thank you!
 

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