Calculating Acceleration of M1,M2 System w/ Friction

[Cglez1280]

Homework Statement

A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between the block and the surface is 0.25. The block is connected by a massless string to the second block with a mass of 300 g. The string passes over a light frictionless pulley. The system is released from rest.
A. Use Newton's second law to write an equation for the 500 g mass
B. Use Newton's second law to write and equation for the 300 g mass
C. Find the acceleration of the system by simultaneously solving the system of two equations.
D. What is the tension force in the string?

M1= 500g
M2= 300g
Coefficient of friction=0.25

Homework Equations

I don't know at the moment

The Attempt at a Solution

So I've answered A and B.
For A I got Ft-Ffr=M1A
For B I got Ft-M2G=M2A
Part C is where I'm completely confused and me help with. I don't really understand how to do that

 

[drvrm]

Part C is where I'm completely confused and me help with. I don't really understand how to do that

you have two equations of motion for the two connected masses.
part C asks you to simultaneously solve for acceleration of the system...
so it must be that the solution should satisfy the two equations...why not substitute result of one in the second so that the second one is also satisfied...
ask a question- should the acceleration of the two masses be same or different?

 

[Cglez1280]

you have two equations of motion for the two connected masses.
part C asks you to simultaneously solve for acceleration of the system...
so it must be that the solution should satisfy the two equations...why not substitute result of one in the second so that the second one is also satisfied...
ask a question- should the acceleration of the two masses be same or different?

The acceleration for the two should be the same I think. This is summer work and my physics teacher gave me the answers and I have to work for the answers. The answer is 2.1 m/s^2

 

[drvrm]

The acceleration for the two should be the same I think.

the same string is connected passingthrough a frictionless pulley so what about the tension in the string?

 

[drvrm]

Check your equations... the equation for the falling mass seems to have an error of sign (directional)!

 

[Cglez1280]

Check your equations... the equation for the falling mass seems to have an error of sign (directional)!

I'm sorry but could you explain? This is my first time ever introduced to any type of physics and well it's hard and this the only question I can't do.

 

[drvrm]

this is daily experience that strings transmit the same tension if it does not extend in length under stress ,
so the force of tension will be same.
the first equation F(tension)- force of friction = mass(A) x Acceleration
the IInd one is Force(gravity)- force(tension) = Mass(B) x Acceleration
so one knows force of friction as coefficient of friction is given. if you use the above two ,you can get the acceleration?

 

[drvrm]

this is daily experience that strings transmit the same tension if it does not extend in length under stress ,
so the force of tension will be same.
the first equation F(tension)- force of friction = mass(A) x Acceleration
the IInd one is Force(gravity)- force(tension) = Mass(B) x Acceleration
so one knows force of friction as coefficient of friction is given. if you use the above two ,you can get the acceleration?

 

[Cglez1280]

this is daily experience that strings transmit the same tension if it does not extend in length under stress ,
so the force of tension will be same.
the first equation F(tension)- force of friction = mass(A) x Acceleration
the IInd one is Force(gravity)- force(tension) = Mass(B) x Acceleration
so one knows force of friction as coefficient of friction is given. if you use the above two ,you can get the acceleration?

Physics is hard oh god. But I guess I'll keep on trying. But from there would you just use substitution to find the acceleration?

 

[drvrm]

But from there would you just use substitution to find the acceleration?

well ,you can add or substract the simultaneous equations ,so that one of the unknown gets eliminated...this is equivalent to substitution, you have two equations so you can find two unknown.

 

[Cglez1280]

well ,you can add or substract the simultaneous equations ,so that one of the unknown gets eliminated...this is equivalent to substitution, you have two equations so you can find two unknown.

Thank you so much I'm pretty sure that I got it now.