Solve Simple Force Question: Acceleration, Tension

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The discussion revolves around solving a physics problem involving a 2.00 kg object on a frictionless table connected to a 7.00 kg hanging object via a pulley. The user derived two equations to find the acceleration and tension in the string, ultimately calculating the acceleration as -7.62. The confusion arose regarding the negative sign of the acceleration, as the user defined upward as negative. Clarification was provided that the negative sign indicates direction based on the chosen coordinate system, and that the magnitude of gravitational acceleration should be treated as positive. The conversation concludes with an acknowledgment of the misunderstanding regarding the signs in physics equations.
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[SOLVED] Simpe Force Question

Homework Statement


A 2.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 7.00 kg object, as in Figure P5.24. Find the magnitude of the acceleration of the two objects and the tension in the string.

The Attempt at a Solution


I have a specific question about this problem that I came upon at the end. Here is how I solved it.

T = m_1a
-T + m_2g = m_2a
So two equations to uknowns, and here is what I got.
-m_1a + m_2g = m_2a
a = \frac{m_2g}{(m_1+m_2)}

So solving for a I got -7.62. The thing I have a question about is the negative on the acceleration. At the beginning of the problem I defined rightward and downward direction as positive and upward direction as negative. So if acceleration is a negative value does that not mean it is traveling up, since I defined up as negative? I know that it doesn't not, but I am confused it can be this way.
 
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Hi Sheneron,

How did you get a negative answer? All the values (masses and g) are positive numbers.

Did you use g=-9.8? The value of g is a magnitude and therefore always a positive number; if we were talking about an object in free fall we would say that its acceleration is

<br /> a_y = -g<br />

just to keep g a positive number.
 
Ah ok I see. thanks
 
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