(a)
<span>Let: </span>
<span>a be the acceleration of the system, </span>
<span>g be the acceleration due to gravity, </span>
<span>T be the tension in the string. </span>
<span>For mA: </span>
<span>T = mA a </span>
<span>For mB: </span>
<span>mB g - T = mB a </span>
<span>Adding to eliminate T: </span>
<span>mB g = (mA + mB)a ...(1) </span>
<span>a = mB g / (mA + mB) </span>
<span>= 7.0 * 9.81 / (11.0 + 7.0) </span>
<span>= 3.815 m/s^2 </span>
<span>= 3.82 m/s^2 to 3 sig. fig. </span>
<span>(b) </span>
<span>Let: </span>
<span>s be the distance to the edge of the table, </span>
<span>t be the time taken. </span>
<span>s = at^2 / 2 </span>
<span>t = sqrt(2s / a) </span>
<span>= sqrt(2 * 1.250 / 3.815) </span>
<span>= 0.810 sec to 3 sig. fig. </span>
<span>(c) </span>
<span>From (1): </span>
<span>(mA + mB)(a / g) = mB </span>
<span>mA(a / g) = mB(1 - a / g) </span>
<span>mA = mB(g / a - 1) </span>
<span>= 4.0(100 - 1) </span>
<span>= 396 kg.</span>
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