A block of mass m1 = 3.56 kg on a horizontal surface is connected to a mass m2 = 21.8 kg that hangs vertically as shown in the figure below. The two blocks are connected by a string of negligible mass passing over a frictionless pulley. The coefficient of kinetic friction between m1 and the horizontal surface is 0.226. Answer both parts to 3 significant figures.

(a) What is the magnitude of the acceleration of the hanging mass? (m/s/s)
(b) Determine the magnitude of the tension in the cord above the hanging mass. (N)

We are given the characteristics of a system where two masses are connected by a massless string. The acceleration under these conditions is common to both objects. By analyzing the acting forces on each mass, we can find both the common acceleration and tension of the string.

Let's analyze the forces acting upon mass m1: To the right, we have the tension of the string that tries to move it in that direction. Opposite to the intent to move is the friction force to the left. Applying the Newton's second law, we have

[tex]T-Fr=m_1.a[/tex]

Where

[tex]Fr=\mu.N=\mu. m_1.g[/tex]

Thus

[tex]T=m_1.a+\mu. m_1.g[/tex]

Now for the mass m2, in the vertical direction

[tex]T-W_2=-m_2.a[/tex]

Note that the sign of the acceleration is downwards since the mass m1 tends to move to the right and both masses are tied. W2 is the weight of the mass 2, thus

[tex]T-m_2.g=-m_2.a[/tex]

Replacing the value of T obtained above

[tex]m_1.a+\mu. m_1.g-m_2.g=-m_2.a[/tex]

Solving for a

[tex]m_1.a+m_2.a=m_2.g-\mu. m_1.g[/tex]

[tex]\displaystyle a=\frac{m_2.g-\mu. m_1.g}{m_1+m_2}[/tex]

[tex]\displaystyle a=\frac{m_2-\mu. m_1}{m_1+m_2}.g[/tex]

Plugging in the given values:

[tex]\displaystyle a=\frac{21.8-0.226\cdot 3.56}{3.56+21.8}.9.8[/tex]

[tex]\boxed{a=8.113\ m/s^2}[/tex]

Computing the tension of the string

[tex]T=3.56\cdot 8.113+0.226\cdot 3.56\cdot 9.8[/tex]

[tex]\boxed{T=36.768\ N}[/tex]

moya1316 moya1316 · Answer 2 · 2021-12-23T14:03:45+01:00

Newton's second law allows us to find the results for the questions about the acceleration and tension of the system, with three significant figures they are:

a) The acceleration of the system is: a = 8.11 m / s²

b) The tension of the rope is; T = 36.8 N

Given parameters

The mass of the blocks m₁ = 3.56kg and m₂ = 21.8 kg

The friction coefficient is μ = 0.226

To find

a) Acceleration.

b) The tension.

Newton's second law states that the net force on a body is proportional to the body's mass and acceleration,

F = m a

Where the bold letters indicate vectors, m is the mass y a is the acceleration of the body.

A free-body diagram is a schematic of the system without the details of the bodies. In the attached we have a free body diagram for the two bodies, we will assume that the vertical direction down is positive.

Let's write Newton's second law.

Body 1

y-axis

N- W₁ = 0

x- axis

T -fr = m₁ a

The friction force is a macroscopic force that represents the interaction between the two surfaces in contact and has the formula

fr = μ N

we substitute

T - μ m₁ g = m₁ a

Body 2

W₂ -T = m₂ a

m₂g - T = m₂ a

Let's write our system of equations.

T = μ m₁ g + m₁ a

-T = -m₂g + m₂a

Let's solve.

(m₁ + m₂) a = ( μ m₁ -m₂) g

[tex]a = \frac{m_2 - \mu \ m_1 }{m_2+m_1 } \ g[/tex]

Let's calculate.

[tex]a = \frac{21.8 - 0.226 \ 3.56}{21.8 + 3.56} \ 9.8[/tex]

a = 8.113 m / s²

Let's substitute in the expression of the hanging block.

m₂g - T = m₂ a

T = m₂ (g-a)

T = 21.8 (9.8- 8.113)

T = 36.78 N

In conclusion, using Newton's second law we can find the results for the questions about the acceleration and tension of the system, with three significant figures they are:

a) The acceleration of the system is: a = 8.11 m / s²

b) The tension of the rope is; T = 36.8 N

Learn more about Newton's second law here: brainly.com/question/25545050