Question

PLEASE HELP!! WORTH A TOTAL OF 100 POINTS!!

The three blocks shown are released from rest and are observed to move at a distance of 0.5m with an acceleration of magnitude of 1.5 m/s^2. answer the following assuming M=2.0kg and ignoring the pulley masses and any friction in the pulleys.

Formulas:
T₁ - m*g = m₁*a --- (1)
T₂ - T₁ - F = m₂*a --- (2)
m₃*g - T₂ = m₃*a --- (3)
m₁ = M
m₂ = 2M
m₃ = 2M

Questions solved:
The magnitude of the friction force: 4.6 N
The coefficient of friction: 0.12
The tension in the left string: 22.6 N
The work done by gravity: 9.8 J
The work done by friction: -2.3 J
The velocity of the blocks after they moved the 0.5 m distance from rest: 1.22 m/s

Needed:
What is the tension in the string on the left?

Answer 1

The last equation gives you the tension in the string on the right:

`T_2 = 2Mg - 2Ma = 2M (g-a) = \boxed{33.2 \, \mathrm N}`