- #1
dancing123
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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!
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!