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

A 0.50 kg skateboard is at rest on a rough, level floor on which two lines have been drawn 1.0 m apart. A constant horizontal force is applied to the skateboard at the beginning of the interval, and is removed at the end. The skateboard takes 8.5 s to travel the 1.0 m distance, and then coast for another 1.25 m before coming to rest. Calculate the force applied to the skateboard, and also the constant frictional force opposing its motion.

*The answers are 2.6 x 10^-2 N and 1.2 x 10^-2 N.

2. Relevant equations

F = ma

a = Δv/Δt

v = d/t

3. The attempt at a solution

v = d/t

v = 2.25m/8.5s

v = 0.2647058824 m/s

a = v/t

a = 0.2647058824m/s /8.5s

a = 0.311418685 m/s^2

F = ma

F = 0.50 kg (0.311418685 m/s^2)

F = 0.0155709343 N

I don't know what I'm doing wrong and it's so fustrating! Also I don't even know how to calculate the frictional force. If anybody could explain and help me out with this I would really appreciate it!

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

2. Relevant equations

3. The attempt at a solution

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# Calculate the force applied to the skateboard, and also the constant frictional force

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