Physics problem finding tension of a string

[nothingatall]

Homework Statement

In Fig. 6-50, block 1 of mass 1.3 kg and block 2 of mass 2.5 kg are connected by a string of negligible mass and are initially held in place. Block 2 is on a frictionless surface tilted at θ = 31o. The coefficient of kinetic friction between block 1 and the horizontal surface is 0.16. The pulley has negligible mass and friction. Once they are released, the blocks move. What then is the tension in the string?

Homework Equations

Maybe T-muFN=m1a
Fn=m1a=12.74
Combined equation: T=-m2a+m2gsin(theta)

The Attempt at a Solution

My T came out to be 4.7N but since it is smaller than my Fn I know it can't be correct.
What equation will give me the correct answer?

 

[PhanthomJay]

Homework Statement

In Fig. 6-50, block 1 of mass 1.3 kg and block 2 of mass 2.5 kg are connected by a string of negligible mass and are initially held in place. Block 2 is on a frictionless surface tilted at θ = 31o. The coefficient of kinetic friction between block 1 and the horizontal surface is 0.16. The pulley has negligible mass and friction. Once they are released, the blocks move. What then is the tension in the string?

Homework Equations

Maybe T-muFN=m1a

Yes, for the block on the horizontal surface

Fn=m1a=12.74

Yes, for the block on the horizontal surface, where a = g in this case.

Combined equation: T=-m2a+m2gsin(theta)

This is the equation for the block on the incline. It is not a combined equation. Use this equation in combination with your first 2 equations to solve for the tension.

The Attempt at a Solution

My T came out to be 4.7N but since it is smaller than my Fn I know it can't be correct.
What equation will give me the correct answer?

Mt T comes out slightly larger, but why do you think T needs to be greater than Fn?

 

[tomcue]

To find the tension in the string, we can use the equation T = m2g(sinθ - μkcosθ), where T is the tension, m2 is the mass of block 2, g is the acceleration due to gravity, θ is the angle of the incline, and μk is the coefficient of kinetic friction. This equation takes into account the effect of friction on the system. Plugging in the given values, we get T = (2.5 kg)(9.8 m/s^2)(sin31° - 0.16cos31°) = 7.2 N. This is the correct tension in the string.