a 5.00kg object placed on a frictionless horizontal table is connected to a string tat passes over a pulley and then is fastened to a hanging 9.00kg object.Find the acceleration of the two objects and the tension in the string.

I mistakenly wrote the equation for the acceleration with two vertically hanging weights on a pulley. With one weight on a frictionless horizontal table, it accelerates faster.

Let M1 = 5 kg and M2 = 9 kg
and T = string tension

T = M1*a
M2*g -T = M2*a
M2*g = (M1 + M2) a
a = M2*g/(M1+M2) = 6.3 m/s^2

T = M1*a = 31.5 N

The two objects accelerate at the same rate. Write and solve two simultaneous F = ma equations, one for each. You should end up with

a = (9-5)/(9+5) * g = 2g/7 = 2.8 m/s^2

T = 5 kg *a = 14.0 N

Student

To find the acceleration of the two objects and the tension in the string, we can use Newton's second law of motion and consider the forces acting on each object individually.

Let's denote the acceleration of both objects as "a" and the tension in the string as "T".

First, consider the 5.00kg object on the table. The only force acting on it is the tension in the string pulling it to the right. According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration (F = ma). Therefore, the equation for the 5.00kg object is:

T = (5.00kg) * a -- Equation 1

Now, consider the 9.00kg object hanging freely. The force of gravity acts downward with a magnitude of (9.00kg) * (9.8m/s^2) = 88.2N. The tension in the string pulls it upward. Since the hanging object is accelerating upward, the net force acting on it is the difference between the tension and the force of gravity:

T - (9.00kg) * (9.8m/s^2) = (9.00kg) * a -- Equation 2

Now we have a system of two equations (Equation 1 and Equation 2) with two unknowns (T and a). We can solve the system of equations simultaneously to find their values.

First, plug Equation 1 into Equation 2:

(5.00kg) * a - (9.00kg) * (9.8m/s^2) = (9.00kg) * a

Simplifying:

(5.00kg - 9.00kg) * a = (9.00kg) * (9.8m/s^2)

-4.00kg * a = 88.2N

Dividing both sides by -4.00kg:

a = -88.2N / 4.00kg

a ≈ -22.05m/s^2

Since the magnitude of acceleration cannot be negative in this context, we consider the positive value:

a ≈ 22.05m/s^2

Now we can substitute the value of acceleration into Equation 1 to find the tension in the string:

T = (5.00kg) * (22.05m/s^2)

T ≈ 110.25N

Therefore, the acceleration of the two objects is approximately 22.05 m/s^2, and the tension in the string is approximately 110.25 N.