MHB Mechanics: Find Coefficient of Friction on 34° Slope w/ 0.4

[Shah 72]

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]

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]

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]

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]

wait ... what?

I tried but iam getting the same ans. I have no clue.

 

[skeeter]

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]

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]

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]

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]

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!