What is the distance moved up the plane (friction problem)

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A particle of mass 3 kg is projected up a plane inclined at 50° with an initial speed of 9 m/s, and the coefficient of friction is 0.5. The normal force is calculated as 19.28 N, resulting in a frictional force of 9.64 N. To find the distance moved up the plane until the particle's velocity reaches zero, the work-energy theorem is suggested, incorporating friction into the calculations. The initial kinetic energy is converted into gravitational potential energy and work done against friction. The correct distance moved up the plane is determined to be 3.72 m, factoring in both gravitational and frictional forces.
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A particle P of mass 3 kg is projected up the line of greatest slope of a plane inclined at an angle of 50° to the horizontal. The coefficient of friction between P and the plane is 0.5. The intiial speed of P is 9 m/s.

ii) Find the frictional force acting on P

ii) What distance moved up the plane by P until its velocity becomes zero.


Homework Equations



f = μ x N where N is the normal

N = m x g x cos(50)


The Attempt at a Solution



i)

N = m x g x cos(50) = 19.28 N

f = 0.5 x 19.28 = 9.64 N


ii) Here I do not know what to do. I thought I could take -5 m/s^2 as a deceleration by using :

f = μ x N

m x a = 0.5 x m x g

a = 5 m/s^2

and then using SUVAT equations but this is wrong since I do not get the answer shown in my book..

Please help!

Thank you!
 
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I did think of that..but shouldn't friction still be taken into account? Because I did the following :

1/2 x m x v^2 = mgh

h = v^2 (2g) = 4.05 where v = 9 m/s and g = 10

and then in order to find the hypotenuse i used

Hypotenuse = 4.05 / sin(50)

But this gives me 5.28 m and the answer is 3.72 m
 
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