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

AI Thread Summary
A 200kg bag of sand is winched up a 10m slope at a 6-degree angle with a coefficient of friction of 0.4, using a 1000N force. The initial speed is 2 m/s, and the goal is to determine the distance moved when the speed decreases to 1.5 m/s. Calculations show that the net force and acceleration lead to a total distance of 37.9m, while the textbook answer is 37.4m. The discussion highlights the importance of correctly applying Newton's laws and considering the impact of friction on motion. Clarification on the slope's length versus height is also suggested for accurate problem-solving.
Shah 72
MHB
Messages
274
Reaction score
0
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
 
Mathematics news on Phys.org
$W_{net} = \Delta KE = \dfrac{1}{2}m(v_f^2-v_0^2) = F_{net} \cdot \Delta x \implies \Delta x = \dfrac{\Delta KE}{F_{net}}$

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

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!
 
Back
Top