Two objects with masses of 4.00 kg and 7.00 kg are connected by a light string t
ID: 1264133 • Letter: T
Question
Two objects with masses of 4.00 kg and 7.00 kg are connected by a light string that passes over a frictionless pulley, as in the figure below.
(a) Determine the tension in the string. (Enter the magnitude only.)
N
(b) Determine the acceleration of each object. (Enter the magnitude only.)
m/s2
(c) Determine the distance each object will move in the first second of motion if both objects start from rest.
m
Explanation / Answer
First, write a force balance on the 4kg object:
The forces are gravity Fg (downward) and tension T (upward), and the object is moving from the problem statement:
m*a = T - mg , or
(4kg)*a = T - (4kg)(9.8m/s^2)
Do the same for the 7 kg object:
m*a = T - mg, or
(7kg)*a = T - (7kg)*(9.8m/s^2)
Since the two blocks are connected via a taut string, a is the same (in magnitude) for both but opposite in sign. We can subtract the two equations to eliminate T from the equations and solve for a:
(4kg + 7kg)*a = (-4kg + 7kg)*(9.8m/s^2)
a= (-4kg + 7kg)*(9.8m/s^2)/(4kg+7kg) = 2.6727 m/s^2 (upward for 4kg block and downward for 7 kg block)
That is the answer to part b)
You can substitute this back into the first equation to solve for T:
T = (4kg)*(2.6727 m/s^2) + (4kg)(9.8m/s^2) = 49.89 N
This is the answer to part a)
For part c) you can use the equation of motion:
d = 1/2*a*t^2
= (0.5)*(2.6727 m/s^2)*(1 s)^2 = 1.336 m (upward for 4kg block, downward for 7kg block)
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