Mass m A rests on a smooth horizontal surface, m B hangs vertically. f m A =15.0
ID: 2138363 • Letter: M
Question
Mass mA rests on a smooth horizontal surface, mB hangs vertically.
f mA=15.0kg and mB=3.0kg in the figure
1)
Mass mA rests on a smooth horizontal surface, mB hangs vertically. f mA=15.0kg and mB=3.0kg in the figure If initially mA is at rest 1.150m 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. 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
(a) mA = 15 kg
mB = 3 kg
The formula for the Tension in the cord, T = mAmBg / (mA+mB) =45**9.8 / 18 = 24.5N
In case of mA, we have T = mA*a
acceleration of mA, a = T / mA = 24.5 / 15 = 1.633 m/s^2
In case of mB, we have mB*a = mB*g - T
a = (mB*g - T) / mB
a = (3*9.8 - 24.5) / 3= 1.633 m/s^2
(b) Initial velocity of mA, u = 0
The distance travelled by mA, s = 1.15 m
We have, s = u t + (1/2) a t^2
s = at^2 / 2
t = (2s / a)^(1/2) = sqrt(2*1.15/ 1.633) = 1.1867sec
(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