So I did an experiment on conservation of momentum, where two carts with different mass start at rest and get accelerated and we measure the speed. What I observed was that after the acceleration, they started to decrease in velocity. I thought this was due to presence of frictional force, and I used regression line to get the "a" of friction. And here's what I got, and from here, I cannot seem to go further because the graph seems like it's contradictory. Shouldn't the slope of both of them be the same as a = ug? (where u is the coefficient of kinetic friction)
$\begingroup$
$\endgroup$
8
-
$\begingroup$ Oh, and Red cart's mass is 0.5037 kg, and Blue cart's mass is 0.2531 kg. $\endgroup$– HellowhatsupOct 12, 2019 at 5:32
-
$\begingroup$ What makes you think that the frictional forces are the same for each cart, the chances are that they are not. Do not try a read more into your data than there is. Analyse the data as found and then comment on your results. $\endgroup$– FarcherOct 12, 2019 at 6:48
-
$\begingroup$ I know that frictional forces are different as Ffriction = ukmg. But at least frictional acceleration should be independent of mass, as A frction = ukg. Also, uk depends on the velocity => higher velocity, higher coefficient of kinetic friction. But the above result just totally contradicts that. $\endgroup$– HellowhatsupOct 13, 2019 at 14:58
-
$\begingroup$ What is $ukg$? What do you mean by frictional acceleration? $\endgroup$– FarcherOct 13, 2019 at 15:10
-
$\begingroup$ So, uk (coefficient of kinetic friction) * g (gravitational acceleration). I meant deaccerelation caused by the friction force. $\endgroup$– HellowhatsupOct 13, 2019 at 15:16
|
Show 3 more comments
2 Answers
$\begingroup$
$\endgroup$
Couple of issues here.
First, assuming these “carts” have wheels we are dealing with rolling resistance (aka rolling friction) not kinetic friction (aka sliding friction). So the force opposing motion is
$$F=C_{cc}N$$
Where $C_{cc}$ is the coefficient of rolling resistance and $N$ is the normal force =$mg$.
Second, unlike the coefficient of kinetic friction, the coefficient of rolling resistance may be a function of the normal force, depending on the nature of the wheel. For pneumatic tires the coefficient may depend on how much tire is in contact with the surface. For instance rolling resistance increases with under inflated tires. Equivalent to an under inflated tire is the same tire supporting more weight.
I bring this up because I notice the magnitude of the coefficient $m$ (not to be confused with m for mass) for your red cart equation is greater than the blue and the red cart weighs twice the blue. Perhaps the coefficient of rolling resistance is higher for the red cart?
Just a thought. Hope it helps.
$\begingroup$
$\endgroup$
The graphs make sense, albeit a bit misleading because of the different scales for red and blue.
Friction opposes motion, so if in $y=b + m t$ if $b$ is positive then $m$ must be negative (blue case) and if $b$ is negative then $m$ must be positive (red case).
The way I read it, we have
$$\begin{array}{l|l} \text{before, t<2.5} & \text{after, t>2.5} \\ \hline v_{\rm blue} = 0 & v_{\rm blue} = v_{\rm blue} = 0.333 - 0.008\,t \\ v_{\rm red} = -0.15 + 0.0156 \,t & v_{\rm red} = ? \end{array}$$
So the $a_{\rm red} = 0.0156$ due to friction is attributed to $$a_{\rm red} = \mu_{\rm red}\,m_{\rm red}g$$ and the $a_{\rm blue} = 0.008$ attributed to $$a_{\rm blue} = \mu_{\rm blue}\,m_{\rm blue} g$$
answered Jul 18, 2020 at 15:14