Solving Physics Homework: Find Acceleration & Tension

[Husker70]

Homework Statement

A 5.00-kg object placed on a frictionless, horizontal table is connected to a cable that
passes over a pulley and then is fastened to a hangng 9.00-kg object. Find the acceleration of the two objects and the tension in the string.

Homework Equations

I am using Net Force = (m2)-1/2(m1)(g)
a = force/total mass

The Attempt at a Solution

I am using Net Force = (m2)-1/2(m1)(g) = (9.00)-1/2(5.00)(9.8) = 63.7N
I used 1/2 of m1 because if the were hanging like on an atwood machine it
would be -m1. Since this is horizontal surface would this be correct?
I then divided the force(63.7N) by Total Mass 14.0-kg to get 4.55m/s2
They both should be travling at the same acceleration, I believe
I'm working on the tension.
Am I starting this right?
Thanks,
Kevin

 

[alphysicist]

Hi Husker70,

Homework Statement

A 5.00-kg object placed on a frictionless, horizontal table is connected to a cable that
passes over a pulley and then is fastened to a hangng 9.00-kg object. Find the acceleration of the two objects and the tension in the string.

Homework Equations

I am using Net Force = (m2)-1/2(m1)(g)
a = force/total mass

The Attempt at a Solution

I am using Net Force = (m2)-1/2(m1)(g) = (9.00)-1/2(5.00)(9.8) = 63.7N
I used 1/2 of m1 because if the were hanging like on an atwood machine it
would be -m1. Since this is horizontal surface would this be correct?

I do not believe this is correct. Rather than try doing this in one step, I would suggest you draw a free body diagram for each of the objects. Then you can write down Newton's law in the vertical direction for the hanging object and in the horizontal direction for the object on the table; with two equations and two unknowns here you can solve for both unknowns. What do you get?

I then divided the force(63.7N) by Total Mass 14.0-kg to get 4.55m/s2
They both should be travling at the same acceleration, I believe

Yes, they will have the same acceleration here.

 

[baba89]

I would like to commend you on your attempt at solving this problem. However, I would like to provide a more comprehensive and accurate solution.

Firstly, the equation you have used, Net Force = (m2)-1/2(m1)(g), is not the correct equation to use in this scenario. This equation is used in the case of an Atwood machine, where the two masses are connected by a string passing over a pulley. In this problem, the two masses are connected by a cable passing over a pulley, which changes the dynamics of the system.

To solve this problem, we need to use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, we have two objects, each with their own mass and acceleration, but they are connected by the cable, so we can consider them as one system.

Let's start by drawing a free-body diagram for each object. The 5.00-kg object on the table has a downward force of 5.00kg x 9.8m/s^2 = 49N due to gravity (assuming a standard value for acceleration due to gravity). Since the table is frictionless, there is no force acting on this object in the horizontal direction. The tension in the cable, T, is acting upwards on the object.

For the hanging 9.00-kg object, we have a downward force of 9.00kg x 9.8m/s^2 = 88.2N due to gravity. The tension in the cable, T, is acting upwards on this object as well.

Now, we can apply Newton's second law to the entire system. The net force acting on the system is equal to the sum of the individual forces acting on each object. In the vertical direction, we have T-T=0, since the tension in the cable is the same for both objects. In the horizontal direction, we have T=ma, where a is the acceleration of the system.

Now, we can set up an equation using these values:
T-T=0
ma-T=0

Solving for T, we get T=ma.

To find the acceleration, we can use the fact that the total mass of the system is 5.00kg + 9.00kg = 14.00kg.