How Does Friction Affect the Final Velocity of a Block on an Incline?

[dancing123]

A block m1 with mass 7 kg moves up an inclined plane with an initial velocity v=4.7 m/s. The inclined plane is at an angle of theta = 45 degrees from the horizontal. The coeff. of kinetic friction between the block and the incline is 0.25.

What is the block's velocity when it has traveled a distance D = 1 m up the incline?



F = (coeff. kinetic friction) x (normal force)
F = ma
vf^2 = vi^2 + 2a(delta x)



This is what I tried:
ma = mu(k) * N
ma = mu(k) * mgsin(theta)
a = mu(k) * gsin(theta)
a = 0.25 * (9.81*sin(45))
a = 1.73

vf^2 = vi^2 + 2a(delta x)
vf= sqrt(vi^2 + 2a(delta x))
vf = sqrt(4.7^2 + (2*1.73*1))
vf =5.05

But this velocity is too large. I don't know what I am doing wrong. Please help!

 

[Disconnected]

Firstly it looks at first glance like you may have taking the frictional force to be positive instead of negative!

By taking both the change in x and the frictional force to be positive, you are implying they are both acting in the same direction, which they clearly are not.

So

v^2=u^2+2ax

Where x=+1, and a=-1.73.

This gives the result

v=SQRT(4.7^2-2*1.73)

Remember you can also do it with conservation of energy, where the change in kinetic energy will be equal to the change in gravitation potential plus the work done against the friction.