So this gives you the mass and force of m1, use Newton's second law (F=m*a) to get the acceleration.... 33.0/4.5kg=7.33 m/s^2
I drew this situation as
[m1]----[m2] since the applied force was to the right the --- tension force becomes limp and should =0
[m1]>~~_[m2] it's easier to imagine pulling a rope, if you take a step inward the rope becomes limp and has no tension force present anymore.
Hope this helps!!
two blocks of masses m1=4.5 kg and m2=6.5 kg resting on a frictionless surface are connected by a light inextensible cord. A horizontal force F of 33.0 N directed to the right is applied to the block with m1 and a horizontal force F of 11N directed to the left is applied to the block with m2. find the acceleration of the system and the tension in the rope.
2 answers
the mass of the system is 11 kg ... 6.5 + 4.5
the net external force is 22 N to the right ... 33 - 11
so the acceleration is ... f / m = 22 / 11 = 2 m/s^2 to the right
the tension in the rope combines with the 11 N force
... to give m2 the 2 m/s^2 acceleration
... t - 11 = 6.5 * 2
the net external force is 22 N to the right ... 33 - 11
so the acceleration is ... f / m = 22 / 11 = 2 m/s^2 to the right
the tension in the rope combines with the 11 N force
... to give m2 the 2 m/s^2 acceleration
... t - 11 = 6.5 * 2