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An Experiment Consists of Pulling a Heavy Wooden Block Across a Level Surface...

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jli10
#1
Jan29-12, 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 free-body 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?
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AdkinsJr
#2
Jan29-12, 04:59 PM
P: 124
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.
jli10
#3
Jan29-12, 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.

Delphi51
#4
Jan29-12, 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 y-intercept (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!
SammyS
#5
Jan29-12, 07:47 PM
Emeritus
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HW Helper
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P: 7,819
Quote Quote by jli10 View Post
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 free-body 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?
The spring force, Fs, was (hopefully) constant during any one trial, but was different from one trial to the next.

The force of friction, Ff, should have been the same across all of the trials.

The data table suggests plotting Fs 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 Fs 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 y-intercept should be the force of friction.

If not, then there was a problem with the experiment.
SammyS
#6
Jan30-12, 09:38 AM
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P: 7,819
The data do fall fairly close to a straight line.


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