# Thank you, I think I get it now. I have one more question:

a kimchi container of mass 1=3.0 kg connected to a block of mass m2 by a cord looped around a frictionless pulley. The cord and pulley have negligible mass. When the container is released from rest, it accelerates at 1.0 m/s^2 across horizontal

Thank you, I think I get it now. I have one more question:

a kimchi container of mass 1=3.0 kg connected to a block of mass m2 by a cord looped around a frictionless pulley. The cord and pulley have negligible mass. When the container is released from rest, it accelerates at 1.0 m/s^2 across horizontal frictionless surface. what are the tension in the cord and mass m2?

I figured it out this way: to figure out he tension and mass I used the equattion T=Mm/M+m(g) and used them interchangebly.

The tension on the m2 is m2*g - m2*a where a is the system acceleration.
That same tension is pulling the 3kg block , so
tension= 3*a

setting these two equal

m2g - m2 *a = 3*a
or 3a +m2a=m2*g
a= m2(g)/(3 +m2)

which makes tension= 3(m2)g/(3+m2) which may be the same as yours, I cant tell without parenthesis in your expression.

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