Mass mA rests on a smooth horizontal surface, mB hangs vertically. Part A If m A
ID: 1277577 • Letter: M
Question
Mass mA rests on a smooth horizontal surface, mB hangs vertically.
Part A
If mA=10.0kg and mB=8.0kg in the figure (Figure 1) , determine the magnitude of the acceleration of each block.
Enter your answers numerically separated by a comma. Express your answers using two significant figures.
Part B
If initially mA 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?
Express your answer using two significant figures.
Part C
If mB=1.0kg, how large must mA be if the acceleration of the system is to be kept at 1100g?
Express your answer using two significant figures.
Explanation / Answer
Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.
Mass mA rests on a smooth horizontal surface, mB hangs vertically.
(a) If mA=11.0 kg and mB=7.0 kg, determine the magnitude of the acceleration of each block.
(b) If initially mA is at rest 1.300 m 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?
(c) If mB=1.0 kg, how large must mA be if the acceleration of the system is to be kept at over 1/100 g?
answer
(a) mA = 11 kg
mB = 7 kg
The formula for the Tension in the cord, T = mAmBg / (mA+mB) = 77*9.8 / 18 = 41.92 N
In case of mA, we have T = mA*a
acceleration of mA, a = T / mA = 41.92 / 11 = 3.81 m/s^2
In case of mB, we have mB*a = mB*g - T
a = (mB*g - T) / mB
a = (7*9.8 - 41.92) / 7 = 3.81 m/s^2
(b) Initial velocity of mA, u = 0
The distance travelled by mA, s = 1.3 m
We have, s = u t + (1/2) a t^2
s = at^2 / 2
t = (2s / a)^(1/2) = (2*1.3 / 3.81) (1/2) = 0.82 sec
(c) mB = 1 kg
The acceleration of the system, a = g/100
The acceleration of the system is given by the formula, a = mB*g / (mA+mB)
mA = [mB*g / a] - mB = [ 1*9.8*100/9.8 ] - 1 = 99 kg