RedAlert Posted February 4, 2007 Share Posted February 4, 2007 A 2 kg mass (m1) and a 3 kg mass (m2) are attached to a lightweight cord that passes over a frictionless pulley. Assume the positive direction of motion to be when the smaller object moves upward and the larger mass moves downward. Okay...I went around trying to solve this as a normal tension problem...going like this: T - (m1)g = (m1)a and T - (m2)g = (m1)a (g = 9.8, a = acceleration) therefore: T = (m1)a + (m1)g and T = (m2)a + (m2)g (m2)a + (m20g = (m1)a + (m2)g solving that gives: a = 9.8 (meters per second squared) the answer key says: 1.96 (meters per second squared) My answer makes more sense to me...but I have a feeling I'm wrong (I get the feeling physics likes to chuck intuition out the window). Is the answer key correct? Link to comment Share on other sites More sharing options...

swansont Posted February 4, 2007 Share Posted February 4, 2007 solving that gives: a = 9.8 (meters per second squared)the answer key says: 1.96 (meters per second squared) My answer makes more sense to me...but I have a feeling I'm wrong (I get the feeling physics likes to chuck intuition out the window). Is the answer key correct? It makes sense to you that the object will fall at g, like an untethered mass, even though the 3 kg mass needs to lift a 2 kg mass? Your intuition needs some honing. You have T - (m1)g = (m1)a T - (m2)g = (m2)a (you had m1, but it should be m2) T and g have opposite signs, which is OK, but a will have a different sign for the two equations. The positive direction was defined to be down for the larger mass, so the second equation is incorrect. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now