# 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!!!