PhysicsMechanics- friction

Shah 72

Member
A particle slides up a slope at angle 34 degree to the horizontal with coefficient of friction 0.4. It passes a point P on the way up the slope with 3m/s and passes it on the way down the slope with speed 2 m/s. Find the coefficient of friction between the particle and the slope.
Coefficient of friction=0.4
F=m×a
While going up
-0.4×10cos 34-10sin 34=a
a=-8.91m/s^2.
V^2=u^2+2as
I get the eq
V^2=9-2×8.91s
I don't understand how to calculate further.
Pls help

skeeter

Well-known member
MHB Math Helper
A particle slides up a slope at angle 34 degree to the horizontal with coefficient of friction 0.4. It passes a point P on the way up the slope with 3m/s and passes it on the way down the slope with speed 2 m/s. Find the coefficient of friction between the particle and the slope.
Coefficient of friction=0.4
F=m×a
While going up
-0.4×10cos 34-10sin 34=a
a=-8.91m/s^2.
V^2=u^2+2as
I get the eq
V^2=9-2×8.91s
I don't understand how to calculate further.
Pls help
wait ... what?

Shah 72

Member
wait ... what?
I don't know how to find the coefficient of friction as if I take v^2= u^2+ 2as, there are two unknown factors which is v and s. So I don't know how to calculate.

topsquark

Well-known member
MHB Math Helper
I don't know how to find the coefficient of friction as if I take v^2= u^2+ 2as, there are two unknown factors which is v and s. So I don't know how to calculate.
skeeter is pointing out that you were given the answer in the problem statement. Look at the bold sections.

-Dan

Shah 72

Member
wait ... what?
I tried but iam getting the same ans. I have no clue.

skeeter

Well-known member
MHB Math Helper
I tried but iam getting the same ans. I have no clue.
Did you read the problem statement? It clearly states the coefficient of friction is 0.4 … why then are you trying to find the coefficient of friction when it’s given?

Shah 72

Member
Did you read the problem statement? It clearly states the coefficient of friction is 0.4 … why then are you trying to find the coefficient of friction when it’s given?
The question asks me to. That's why this question is so confusing. And the ans in the textbook is 0.259

topsquark

Well-known member
MHB Math Helper
The question asks me to. That's why this question is so confusing. And the ans in the textbook is 0.259
Then something is wrong with the problem statement. Talk to your instructor.

-Dan

skeeter

Well-known member
MHB Math Helper
up the incline …

$a = -[10\sin(34)+4\cos(34)] = -8.91 \, m/s^2$

$\Delta x = \dfrac{0^2-3^2}{2a} = 0.51 \, m$

down the incline …

$a = \dfrac{2^2-0^2}{2 \Delta x} = 3.96 \, m/s^2$

$\mu = \dfrac{10\sin(34) - a}{10\cos(34)} = 0.2$

not a very well engineered problem, imo.

Shah 72

Member
up the incline …

$a = -[10\sin(34)+4\cos(34)] = -8.91 \, m/s^2$

$\Delta x = \dfrac{0^2-3^2}{2a} = 0.51 \, m$

down the incline …

$a = \dfrac{2^2-0^2}{2 \Delta x} = 3.96 \, m/s^2$

$\mu = \dfrac{10\sin(34) - a}{10\cos(34)} = 0.2$

not a very well engineered problem, imo.
Tha
up the incline …

$a = -[10\sin(34)+4\cos(34)] = -8.91 \, m/s^2$

$\Delta x = \dfrac{0^2-3^2}{2a} = 0.51 \, m$

down the incline …

$a = \dfrac{2^2-0^2}{2 \Delta x} = 3.96 \, m/s^2$

$\mu = \dfrac{10\sin(34) - a}{10\cos(34)} = 0.2$

not a very well engineered problem, imo.
Thank you very very much!!!