Two masses on an inclined plane with a massless pulley

AI Thread Summary
The discussion revolves around a physics problem involving two masses connected by a string over a pulley, with one mass on a 20-degree incline and the other hanging. The minimum mass required to prevent slipping on the incline was calculated to be 1.829 kg. However, the user encountered issues calculating the acceleration after nudging the mass, consistently arriving at an incorrect value of 2.76 m/s². The solution involves applying Newton's second law to both masses, incorporating tension and friction forces to create two equations with two unknowns. Properly setting up these equations will lead to the correct calculation of acceleration.
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Homework Statement


A block of mass m resting on a 20 degree slope. The block has coefficients of friction mu_s =0.80 and mu_k =0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

What is the minimum mass m that will stick and not slip?

If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Homework Equations


T-f_k-(mg*sin(theta))=m*a
f_k=u_k*(mg*cos(theta))

The Attempt at a Solution


I completed the first part of the problem and determined that the minimum mass is 1.829 kg. I calculated the tension on the rope by multiplying the mass of the hanging block by 9.8. But when I plug these values into the equation I keep getting 2.76 m/s^2 which is not the correct answer. Is there something wrong with my equations or the tension value?
 
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The tension in the rope is only equal to the weight of the hanging block if the objects aren't accelerating. You need to write the sum of the forces = ma for each object, where the sum of the forces includes the unknown tension T. Then you have two equations and two unknowns and you solve for T and a.
 
Do you mean like T-(u_k*(mg*cos(theta)))-(mg*sin(theta))=m*a?
 
I mean something exactly like that.
 
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