
#1
Jan2912, 03:33 PM

P: 6

1. The problem statement, all variables and given/known data
An experiment consists of pulling a heavy wooden block across a level surface with a spring force meter. The constant force for each try is recorded, as is the acceleration of the block. The data are shown below: Force F in Newtons: 3.05  3.45  4.05  4.45  5.05 acceleration a in meters/second^2: 0.095  0.205  0.295  0.405  0.495 (The above is supposed to be a table where every force corresponds with a different acceleration.) 2. Relevant equations ƩF=ma F_f(friction force)=μN 3. The attempt at a solution So I don't really what the constant force is here. Is it the spring force, and if it is, should I use the formula F=kx? Well, I drew the freebody diagram and I got a couple of things. In the vertical direction, there is no acceleration so the normal force N is equal to mg. In the horizontal direction, I got: [itex] \sum{F} = ma = F_s  F_f = F_s  (\mu)(N) = F_s  (\mu)(mg) \Rightarrow a = \frac{F_s}{m}  (\mu)(g). [/itex] Where do I go from here, and more specifically, how do I incorporate the table values into my solution? 



#2
Jan2912, 04:59 PM

P: 122

The spring force is a variable force, not a constant. If you plot force vs. displacement for a spring force it is a line F=kx where x is the displacment from equilibrium and k is the slope (spring constant).
The data you present also shows increasing force. It's also not clear what the object of the problem is. 



#3
Jan2912, 05:39 PM

P: 6

Sorry, I forgot to include the question statement:
Which is the best value for the mass of the block? a) 3 kg b) 5 kg c) 10 kg d) 20 kg e) 30 kg Thanks for your reply AdkinsJr. So if the spring force is not the specified constant force in the problem, what exactly is? Also, if the force is supposedly "constant," then why are the values in the table increasing? I'm so lost right now... Any further help would be appreciated. 



#4
Jan2912, 06:38 PM

HW Helper
P: 3,394

An Experiment Consists of Pulling a Heavy Wooden Block Across a Level Surface...
Graph force vs acceleration. Make sure you start at 0 on both axes when you draw the graph. In experiments, you always hope for a straight line  you have one here. So you can write an equation for it easily. The equation for a straight line is y = mx+b, or so they teach in grade 9 math where you always graph y vs x. Instead of x, you have "a" for acceleration. Instead of y, you have . . .
Make those changes in your equation. The "m" in the formula represents slope; you can get the number for that off your graph. The "b" represents the yintercept (where the line hits the vertical axes, so you can read that number off your graph. After the experimental work is done, consider the theory. Do you have any formulas relating F and a? Good old F = ma should apply to this situation but it is like y = mx with a zero vertical intercept so it doesn't quite work. It is as if you have a little something extra added to the ma or taken away from the F in this case. That something extra has units of force. Any forces involved here other than the one you measure with the spring scale? The ma is actually the net force that causes the acceleration  what is left after all forces are combined. See what you come up with! 



#5
Jan2912, 07:47 PM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,398

The force of friction, F_{f}, should have been the same across all of the trials. The data table suggests plotting F_{s} versus acceleration, a. Rewriting your equation, [itex]\displaystyle ma = F_s  F_f[/itex] as [itex]\displaystyle F_s = ma + F_f[/itex]I suggest putting F_{s} on the vertical axis, and a on the horizontal axis. Hopefully, your data will nearly fall along a straight line. If so, the the slope should be the mass of the block, and the yintercept should be the force of friction. If not, then there was a problem with the experiment. 



#6
Jan3012, 09:38 AM

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P: 7,398

The data do fall fairly close to a straight line.



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