Respuesta :

(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.

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