Calculate Skateboard Force & Friction: 2.6x10^-2 & 1.2x10^-2N

[plumeria28]

Homework Statement

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.

Homework Equations

F = ma

a = Δv/Δt

v = d/t

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!

 

[Delphi51]

Welcome to PF!
Wow, a neat question - compliments to your prof.

v = d/t
v = 2.25m/8.5s

It is accelerated motion so you can't use v = d/t. Use d = Vi*t + .5*at² for starters.
I think you will need more formulas; keep your list of constant acceleration formulas handy!

 

[Spinnor]

What you don't know,

F applied force
f constant frictional force
a constant acceleration over first interval
d constant deceleration over second interval
v_o velocity at end of 1m interval
t2 time to go the second interval

You know,

t1 time to go the first interval = 8.5s
F - f = ma acceleration for first interval
f = md acceleration for second interval
v = v_o = at1 velocity for the first interval
v = v_o - dt2 velocity for the second interval
x = at1^2/2 = 1m for the first interval
x = v_ot2 - dt2^2/2 = 1.25m for the second interval

Did I leave out anything? I think you then have six equations in six unknowns?

 

[netgypsy]

Always remember that F = ma is the SUM of the forces on that particular axis = ma

You have enough information for the first interval to calculate the overall acceleration with both forces applied using your x = 1/2atsquared
Find your net force using Newton's second law.

and your final velocity using Fx = 1/2mvsquared (work = change in kinetic energy)

Using the final velocity from part one and the fact that it is at rest at the end plus the given distance you can use the same work energy relation to find find the frictional force, then subtract it from the total force found in part a to determine the force pulling it.

 

[plumeria28]

Thanks for the help everyone! :)

 

[netgypsy]

at your service :-)