Question

The figure below shows two objects connected to each other by a light cord that passes over a pulley with negligible

friction. The object of mass m₂ = 4.00 kg lies on a horizontal table with negligible friction, while the object of mass
m₂ 11.7 kg hangs vertically.
(a) What is the magnitude of the acceleration of each object (in m/s2)? (Here, a, is the acceleration of m₂, and a, is the
acceleration of m₂.)
7.3
7.3
3₂ -
✓ m/s²
m/s²
(b) What is the tension in the cord (in N)?
4.9
x
Apply Newton's second law to each object separately, drawing free-body diagrams for both. How are the accelerations
of the two objects related to each other? How is the tension force on each related to the other? Can you set up
equations that allow you to solve simultaneously for the acceleration and tension? N

Answer 1

To solve this problem, we can apply Newton's second law to each object separately and set up equations that relate the accelerations and tension.

Let's analyze each object individually:

Object m₁ (mass = m₁):

It is on a horizontal table with negligible friction.

The only force acting on m₁ is the tension in the cord, T.

Object m₂ (mass = m₂):

It is hanging vertically.

The only force acting on m₂ is its weight, which can be calculated as m₂ * g, where g is the acceleration due to gravity.

Now, let's set up the equations:

For object m₁:

T = m₁ * a₁ -- Equation (1) (acceleration of m₁)

For object m₂:

m₂ * g - T = m₂ * a₂ -- Equation (2) (acceleration of m₂)

The acceleration of m₁ is in the same direction as the tension T, while the acceleration of m₂ is opposite to the tension T.

To solve for the acceleration and tension, we need to eliminate one variable. We can do this by substituting Equation (1) into Equation (2):

m₂ * g - m₁ * a₁ = m₂ * a₂

Now, we can solve for the acceleration a₂:

a₂ = (m₂ * g - m₁ * a₁) / m₂ -- Equation (3)

To find the tension, we can substitute the value of a₂ from Equation (3) into Equation (1):

T = m₁ * a₁

Now, we can solve for the acceleration a₁:

a₁ = T / m₁ -- Equation (4)

Finally, we can substitute the value of a₁ from Equation (4) into Equation (3) to find a₂:

a₂ = (m₂ * g - m₁ * (T / m₁)) / m₂

a₂ = (m₂ * g - T) / m₂

Now, we have two equations that allow us to solve simultaneously for the acceleration (a₁ and a₂) and tension (T).

Given that a₁ = 7.3 m/s², we can substitute this value into Equation (4) to find T:

T = m₁ * a₁

T = m₁ * 7.3

To find the tension in the cord, we need the value of m₁. Please provide the mass of object m₁, and we can calculate the tension and the acceleration of m₂ accordingly.