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avsj
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Homework Statement
A 9.0 kg box is sliding down a smooth flat ramp which makes an angle of 24 degrees with the horizontal. If the coefficient of kinetic friction (μ) between the box and the ramp is 0.25, how long will it take the box to slide 2.0 m down the ramp froma standing start?
Homework Equations
Fg = mg
Ff = μFn
SOH/CAH/TOA
The Attempt at a Solution
Fg= 9.0 x 9.8 =88.2 N
From a vector diagram, I solved Fn using cos(24)x88.2= 80.6 N
Therefore, Ff = 0.24 (80.6) = 20.2 N
Fd (force down the ramp parallel to ramp) = sin(24) x 88.2 = 35.9
Fnet = 35.9 - 20.2 = 15.7 N
Now to solve d=vt.
d= 2.0m
but do i use 15.7 as my velocity or is it acceleration and if so, how should I approach the calculus? Or is the velocity also 15.7 m/s because it starts at rest. If so, i divide 2 by 15.7 and get a 0.13 but the answer in my textbook is 1.5
Thanks for your help