A 1.0-kg mass (mA) and a 7.0-kg mass (mB) are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. Find the acceleration of the larger mass

a = [(mB-mA)/(mB+mA)]*g = (3/4)g

You can derive it by writing free-body motion equations for each mass, and eliminating the cord tension force, T.

mA*a = T - mA*g
mB*a = mB*g -T

(mB + mA)*a = (mB - mA)*g

To find the acceleration of the larger mass (mB), we can use Newton's second law of motion which states that the acceleration of an object is directly proportional to the net force applied on it and inversely proportional to its mass.

Let's denote the acceleration as "a" and the net force as "F_net".

1. Identify the forces acting on the larger mass (mB):
- The force due to the weight of the larger mass (mB*g), where g is the acceleration due to gravity.
- The tension force in the lightweight cord (T), which is the same on both masses due to their connection.

2. Write down the equations of motion for both masses:
For mA:
- T - mA*g = mA*a (equation 1)

For mB:
- mB*g - T = mB*a (equation 2)

3. Since T is the same in both equations, let's solve equation 1 for T and substitute it into equation 2:

T = mA*a + mA*g (substituting T in equation 2)

mB*g - mA*a - mA*g = mB*a

4. Rearrange the equation to isolate the acceleration (a):

mB*g - mA*g = mB*a + mA*a

(mB - mA)*g = (mB + mA)*a

a = [(mB - mA)*g] / (mB + mA)

5. Plug in the known values:
- mB = 7.0 kg (mass of the larger mass)
- mA = 1.0 kg (mass of the smaller mass)
- g = 9.8 m/s^2 (acceleration due to gravity)

a = [(7.0 kg - 1.0 kg) * 9.8 m/s^2] / (7.0 kg + 1.0 kg)

a = (6.0 kg * 9.8 m/s^2) / 8.0 kg

a = 58.8 m/s^2 / 8.0 kg

a = 7.35 m/s^2

The acceleration of the larger mass (mB) is 7.35 m/s^2.

To find the acceleration of the larger mass (mB), we can use Newton's second law of motion which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, the force causing the acceleration of the system is the tension in the cord. Assuming the cord has negligible mass, the tension in the cord will be the same on both sides of the pulley.

Let's denote the acceleration of the system as 'a' and the tension in the cord as 'T'.

For mass mA:
The force acting on mass mA is its weight, which is given by:
FmA = mAg
where m is the mass of mA and g is the acceleration due to gravity (approximately 9.8 m/s²).

For mass mB:
The force acting on mass mB is the tension in the cord, but this force acts in the opposite direction, so we can write it as:
FmB = TB - mBg
where TB is the tension in the cord on the side of mass mB.

Since the cord is connected, the tension TB is the same as the tension TA on the side of mass mA.

Equating the two forces and expressing the tensions in terms of mB and mA:
mAg = mBg - (TB - mBg)
mAg = mBg - TB + mBg
mAg = 2mBg - TB
TB = 2mBg - mAg

Substituting the value of TB in the equation for FmB:
FmB = 2mBg - mAg - mBg
FmB = mBg - mAg

Now, according to Newton's second law, the force is equal to the mass times the acceleration:
FmB = mBa

Substituting the value of FmB:
mBg - mAg = mBa

Rearranging the equation to solve for the acceleration:
a = (mBg - mAg) / mB

Plugging in the values:
a = [(7.0 kg)(9.8 m/s²) - (1.0 kg)(9.8 m/s²)] / 7.0 kg

Simplifying the equation:
a = (68.6 N - 9.8 N) / 7.0 kg
a = 58.8 N / 7.0 kg

Calculating the acceleration:
a ≈ 8.4 m/s²

Therefore, the acceleration of the larger mass (mB) is approximately 8.4 m/s².

9 m/s2