PhysicsMechanics- friction

Shah 72

Member
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

skeeter

Well-known member
MHB Math Helper
$W_{net} = \Delta KE = \dfrac{1}{2}m(v_f^2-v_0^2) = F_{net} \cdot \Delta x \implies \Delta x = \dfrac{\Delta KE}{F_{net}}$

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

I agree with the text solution ...

Shah 72

Member
$W_{net} = \Delta KE = \dfrac{1}{2}m(v_f^2-v_0^2) = F_{net} \cdot \Delta x \implies \Delta x = \dfrac{\Delta KE}{F_{net}}$

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

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!

skeeter

Well-known member
MHB Math Helper
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

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

try again ...

Shah 72

Member
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

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

try again ...
Sure I will try. Thanks a lotttt!

Shah 72

Member
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

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

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.

Shah 72

Member
$a = 5 - g(\sin{\theta} + \mu \cos{\theta})$

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

try again ...
I tried and I got the ans. Thank you!

skeeter

Well-known member
MHB Math Helper
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.

Shah 72

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