- #1
gelfand
- 40
- 3
Homework Statement
A block of mass 100 grammes was stationary on the flat surface at ##x = 0##. At
time ##t = 0## a horizontal force of 10 Newtons was applied on the block in the
positive x direction during ##\Delta t = 3## seconds.
Find at what x position the block will stop, if the coefficient of kinetic
friction between the block and the surface is ##\mu_k = 0.2##.
Homework Equations
##I = F \Delta t = m \Delta v = \Delta p## (where ##p## is used for momentum)
The Attempt at a Solution
I have that$$
F = ma
$$
Then
$$
\frac{10}{0.1} = a
$$And I can presume (?) that acceleration is constant.Then from this I have
$$
a = 100
$$
So if I have ##a = 100## then I can use Impulse as ##I = F \Delta t## to find
$$
F \Delta t = \Delta p
$$
So
$$
(0.1) (100) (3) = \Delta p = 30
$$So from ##\Delta p = 30## I have ##m(v_f - v_0) = 30## and as ##v_0 = 0## this is
##mv_f = 30## and ##v_f = \frac{30}{0.1} = 300##
This gives me the velocity as ##300## m/s at the point the force has been
removed.
From here I need to know how long it takes for the force of friction to slow the
object to a stop.
This means all of the energy from the point ##t = 3## is spent on friction.
So
$$
\frac{1}{2} mv^2 = \text{work done against friction}
$$So here I have that work done is ##W = F \Delta x## so given ##\mu_k = 0.2##, the
mass of the object is ##0.1## then I would have the force of friction as
$$
F_{friction} = mg \times \mu_k = 0.1 \times 9.81 \times 0.2 = 10.11
$$So then I have
$$
\frac{1}{2} (0.1) v^2 = 10.11 \Delta x
$$
And so
$$
\frac{1}{2} (0.1) v^2 \times \frac{1}{10.11} = \Delta x
$$
Using the velocity found from before this gives $$
\frac{1}{2} (0.1) (300)^2 \times \frac{1}{10.11} = \Delta x
$$
So ##\Delta x \approx 445.103## meters.
I don't really have much of a handle on what to expect from an answer here.
I think what I've done is reasonable though?
Thanks