- Thread starter
- #1

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

- Thread starter Shah 72
- Start date

- Thread starter
- #1

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

- Mar 1, 2012

- 935

A particle slides up a slope at angle 34 degree to the horizontalwith 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

- Thread starter
- #3

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.wait ... what?

- Aug 30, 2012

- 1,207

skeeter is pointing out that you were given the answer in the problem statement. Look at the bold sections.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.

-Dan

- Thread starter
- #5

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

- Mar 1, 2012

- 935

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

- Thread starter
- #7

The question asks me to. That's why this question is so confusing. And the ans in the textbook is 0.259Did 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?

- Aug 30, 2012

- 1,207

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

-Dan

- Mar 1, 2012

- 935

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

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

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.

- Thread starter
- #10

Tha

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

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

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!!!

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

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

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.