# A 4kg mass sits on a frictionless table....

• MrBeans
In summary, the conversation is discussing the calculation of the net force on a 4kg mass with a 20N force applied and a tension force from a string attached to a 2kg mass. The correct calculation takes into account both forces and the sum of the two masses, resulting in a different value than initially calculated. The importance of drawing a free body diagram and using correct order of operations is emphasized.
MrBeans
Homework Statement
A 4kg mass sits on a frictionless table. A string is attached and leads to the right off the table, over a pulley where it then supports a 2 kg mass. The 4 kg mass is being pulled to the left along the table by a 20 N force. What is the tension in the connecting string?
A. 13N
B. 19.7N
C. 6.4N
D. 35N
Relevant Equations
F = ma
T = mg - ma
I got 13N but is that right because apparently, it's wrong
Here's my work:
F = mg = 2(10) = 20N
F = ma
a = F/m = 20/4+2 = 20/6 = 10/3 = 3.3m/s^2
T = mg - ma T = (2kg)(10m/s^2) - (2kg)(3.3m/s^2) = 13.4 N

I appreciate it! And if I'm wrong could you show how you got your answer? Thanks

MrBeans said:
Homework Statement:: A 4kg mass sits on a frictionless table. A string is attached and leads to the right off the table, over a pulley where it then supports a 2 kg mass. The 4 kg mass is being pulled to the left along the table by a 20 N force.
Let me do this quickly in my head. Indeed, I get something different. So let us check your work.

As I understand it, we have two forces on the 4 kg mass. One of 20 N is applied directly with no mechanism specified. The other from the tension in the string.

MrBeans said:
F = mg = 2(10) = 20N
So here, F is the force of gravity on the dangling 2 kg mass.

MrBeans said:
F = ma
But here, F is the net force on... something.

It is wise to avoid using the same variable name in the same problem to denote two different quantities.

MrBeans said:
a = F/m = 20/4+2 = 20/6 = 10/3 = 3.3m/s^2
Apparently that something has a mass of 4+2. So it is the sum of the two masses. But you only counted one of the two forces.

Note that you wrote 20/4+2 where you presumably meant 20/(4+2) = ##\frac{20}{4+2}##. Order of operations is important.

topsquark, Lnewqban and PeroK
Because you did not draw a free body diagram for each mass you have neglected a force. Do it correctly please.

topsquark and jbriggs444
jbriggs444 said:
Let me do this quickly in my head. Indeed, I get something different. So let us check your work.

As I understand it, we have two forces on the 4 kg mass. One of 20 N is applied directly with no mechanism specified. The other from the tension in the string.So here, F is the force of gravity on the dangling 2 kg mass.But here, F is the net force on... something.

It is wise to avoid using the same variable name in the same problem to denote two different quantities.Apparently that something has a mass of 4+2. So it is the sum of the two masses. But you only counted one of the two forces.

Note that you wrote 20/4+2 where you presumably meant 20/(4+2) = ##\frac{20}{4+2}##. Order of operations is important.
oh, I got it. Thank you thank you.

## 1. What is the acceleration of the mass?

The acceleration of the mass is zero, since there is no external force acting on it and there is no friction to oppose its motion.

## 2. What is the net force acting on the mass?

The net force acting on the mass is also zero, since there are no external forces acting on it and there is no friction to oppose its motion.

## 3. Can the mass move on its own?

No, the mass cannot move on its own since there is no external force acting on it and there is no friction to initiate its motion.

## 4. What would happen if the table was not frictionless?

If the table was not frictionless, the mass would experience a frictional force that would oppose its motion and cause it to slow down and eventually come to a stop.

## 5. Is this scenario realistic?

No, this scenario is not realistic since there is always some amount of friction present in real-world situations. However, it is a common theoretical scenario used in physics to simplify calculations and understand fundamental concepts.

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