- #561
lawtonfogle
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Taken [tex]\sum F = m_1 \cdot a [/tex]
We have [tex] \sum F_x = m_1 \cdot a_x [/tex] and [tex] \sum F_y = m_1 \cdot a_y [/tex]
With [tex] a_y = 0 [/tex]
We We have [tex] \sum F_x = m_1 \cdot a_x [/tex] and [tex] \sum F_y = 0 [/tex]
So [tex] T = m_1 \cdot a_x [/tex]
Taken [tex]\sum F = m_2 \cdot a [/tex]
We have [tex]\sum F_y = m_2 \cdot a_y [/tex]
Which is [tex]m_2 \cdot g - T = m_2 \cdot a [/tex]
Inserting [tex] T = m_1 \cdot a_x [/tex] into [tex]m_2 \cdot g - T = m_2 \cdot a [/tex] and solving for [tex]a[/tex],
We get [tex] a = \frac {m_2 \cdot g} {m_1 + m_2}[/tex]
Inserting this into [tex] T = m_1 \cdot a[/tex],
we get [tex] T = \frac {m_1 \cdot m_2 \cdot g} {m_1 + m_2} [/tex]
Now, taken [tex] v^2 = v_0^2 + 2a\Delta x [/tex]
Solving for [tex] a [/tex] we get [tex] a = \frac {v^2 - V_0^2} {2\Delta x}[/tex]
Taking that [tex] \Delta x = .5 [/tex] and [tex] v_0 = 0 [/tex]
We have [tex] a = \frac {v^2} {meters*} [/tex]
Now [tex] a_exp [/tex] stands for experimental value of [tex] a [/tex]
and [tex] a_t [/tex[ stands for theoretical value of [tex] a [/tex]
so [tex] a_exp = \frac {v^2} {meters*} [/tex]
and [tex] a_t = \frac {m_2 \cdot g} {m_1 + m_2}[/tex]
* meters is added so units sovle correctly.
We have [tex] \sum F_x = m_1 \cdot a_x [/tex] and [tex] \sum F_y = m_1 \cdot a_y [/tex]
With [tex] a_y = 0 [/tex]
We We have [tex] \sum F_x = m_1 \cdot a_x [/tex] and [tex] \sum F_y = 0 [/tex]
So [tex] T = m_1 \cdot a_x [/tex]
Taken [tex]\sum F = m_2 \cdot a [/tex]
We have [tex]\sum F_y = m_2 \cdot a_y [/tex]
Which is [tex]m_2 \cdot g - T = m_2 \cdot a [/tex]
Inserting [tex] T = m_1 \cdot a_x [/tex] into [tex]m_2 \cdot g - T = m_2 \cdot a [/tex] and solving for [tex]a[/tex],
We get [tex] a = \frac {m_2 \cdot g} {m_1 + m_2}[/tex]
Inserting this into [tex] T = m_1 \cdot a[/tex],
we get [tex] T = \frac {m_1 \cdot m_2 \cdot g} {m_1 + m_2} [/tex]
Now, taken [tex] v^2 = v_0^2 + 2a\Delta x [/tex]
Solving for [tex] a [/tex] we get [tex] a = \frac {v^2 - V_0^2} {2\Delta x}[/tex]
Taking that [tex] \Delta x = .5 [/tex] and [tex] v_0 = 0 [/tex]
We have [tex] a = \frac {v^2} {meters*} [/tex]
Now [tex] a_exp [/tex] stands for experimental value of [tex] a [/tex]
and [tex] a_t [/tex[ stands for theoretical value of [tex] a [/tex]
so [tex] a_exp = \frac {v^2} {meters*} [/tex]
and [tex] a_t = \frac {m_2 \cdot g} {m_1 + m_2}[/tex]
* meters is added so units sovle correctly.