(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Fg = mg

Ff = μFn

SOH/CAH/TOA

3. 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 text book is 1.5

Thanks for your help

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Vector ramp question - time taken for object to slide down ramp

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