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