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Physics Mechanics- friction

Shah 72

Member
Apr 14, 2021
193
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
Mar 1, 2012
935
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
Apr 14, 2021
193
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
Aug 30, 2012
1,207
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
Apr 14, 2021
193

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
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
Apr 14, 2021
193
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
Aug 30, 2012
1,207
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
Mar 1, 2012
935
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
Apr 14, 2021
193
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!!!