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1. We know from Newton’s second law that F =
ma. If we decided to plot ln(F) on the y-axis and ln(m) on the x-axis, what would we
find for our slope? |
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2. Considering the ln(F)
– ln(m) plot in question (1), what would we find the y-intercept to be? |
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3. Alternatively, one might plot Force (F)
on the y-axis and mass (m) on the x-axis, a best straight line of this plot gives
the equation, F = 2.30 (N/kg) m + 1.4 N.
What was the constant acceleration used in collecting the data that
resulted in this equation? |
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4. What can you say about the y-intercept of
1.4 N? |
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Since F=ma should
have no y-intercept term, the 1.4 N is a measure of error in the system. It could come from a variety of sources,
friction between the cart and the track, tension not being exactly
horizontal, error measuring the flag width which would then result in
inaccurate velocities which could produce inaccurate accelerations, etc. |
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OVER à |
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5. Assuming the track was frictionless, draw
a free body diagram for the cart mass and the hanging mass independently. |
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6. Write out the three relevant sum of the
forces equations. |
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7. From these equations determine the acceleration of the masses and the tension in the
string. |
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Use |
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to substitute
for T in |
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Rearrange and
solve for a. |
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Now with a plug
that back in to find T |
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