Two objects with masses of 2.10 kg and 5.60 kg are connected by a light string t

Question

Two objects with masses of 2.10 kg and 5.60 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.) (b) Determine the acceleration of each object. (Enter the magnitude only.) (c) Determine the distance each object will move in the first second of motion if both objects start from rest.

(a) Determine the tension in the string. (Enter the magnitude only.) (b) Determine the acceleration of each object. (Enter the magnitude only.)

(c) Determine the distance each object will move in the first second of motion if both objects start from rest.

Explanation / Answer

First, write a force balance on the 3kg 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 (2.10kg)*a = T - (2.10kg)(9.8m/s^2) Do the same for the 5.10 kg object: m*a = T - mg, or (5.10kg)*a = T - (5.10kg)*(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: (2.10kg + 5.10kg)*a = (-2.10kg + 5.10kg)*(9.8m/s^2) a= (-2.10kg + 5.10kg)*(9.8m/s^2)/(2.10kg+5.10kg) = 3.92 m/s^2 (upward for 3kg 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 = (2.10kg)*(3.92m/s^2) + (2.10kg)(9.8m/s^2) = 41.15N 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)*(3.92m/s^2)*(1 s)^2 = 1.96m (upward for 3kg block, downward for 7kg block) Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.