Mass mA rests on a smooth horizontal surface, mB hangs vertically. Part A If mA=
ID: 2053001 • Letter: M
Question
Mass mA rests on a smooth horizontal surface, mB hangs vertically.
Part A
If mA=11.0kg and mB=7.0kg in the figure , determine the magnitude of the acceleration of each block. m/s^2
Part B
If initially ms is at rest 1.200m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely? s
Part C
If mB=1.0kg, how large must mA be if the acceleration of the system is to be kept at 1/100g? kg
Figure url ->(http://session.masteringphysics.com/problemAsset/1057072/8/GIANCOLI.ch04.p52.jpg)
Explanation / Answer
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 * 1.200 / 3.815)
=sqrt(0.62 ) sec
(c)
From (1):
(mA + mB)(a / g) = mB
mA(a / g) = mB(1 - a / g)
mA = mB(g / a - 1)
= 1.0(100 - 1)
= 99 kg.