A block with mass m1 = 9.4 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of μk = 0.52. A second block with a mass m2 = 11.5 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. The second block is released and motion occurs.

a) Write an expression for the magnitude of acceleration of block 2, a2, in terms of the acceleration of block 1, a1. (Assume the cable connecting the masses is ideal)
b) Write an expression using the variable provided for the magnitude of the tension force, T.
c) What is the tension, T, in newtons?

For block m1, the one which is on the horizontal table, there are two forces: the tension of the string, T, and the force of friction, [tex]\mu_k m_1 g[/tex], pulling in the opposite direction. So the equation of motion is

[tex]T-\mu_k m_1 g = m_1 a_1[/tex] (1)

For block m2, the one which is hanging vertically, there are two forces: the weight [tex]m_2 g[/tex], pushing downward, and the tension in the cable, acting upward. So the equation of the forces is

[tex]m_2 g - T = m_2 a_2[/tex] (2)

(note that if the cable is inextensible, the two blocks have same acceleration, so [tex]a_1 = a_2[/tex]).

We can isolate the tension T from eq.(1):

[tex]T=m_1 (a_1 +\mu_k g)[/tex]

And substitute it into eq. (2):

[tex]m_2 g - m_1 (a_1 +\mu_k g) = m_2 a_2\\a_2 =g-\frac{m_1}{m_2}(a_1 + \mu_k g)[/tex]

b) [tex]T=\frac{m_1 m_2g (1+\mu_k)}{m_1 + m_2}[/tex]

Since the accelerations of the two blocks are equal: [tex]a_1 = a_2=a[/tex], we can rewrite the two equations of motion as

[tex]T=m_1 (a +\mu_k g)[/tex] (1)

[tex]a=g-\frac{T}{m_2}[/tex] (2)

So we can substitute the expression for a from (2) into (1) and solve the equation for T, the tension:

[tex]T=m_1 (g-\frac{T}{m_2})+\mu_kg)\\m_2 T=m_1 m_2 g - m_1 T + m_1 m_2 \mu_k g\\T=\frac{m_1 m_2g (1+\mu_k)}{m_1 + m_2}[/tex]

c) 77.0 N

In order to find the magnitude of the tension, we just need to substitute data into the equation find at point b):

[tex]T=\frac{(9.4 kg)(11.5 kg)(9.8 m/s^2) (1+0.52)}{9.4 kg + 11.5 kg}=77.0 N[/tex]

nutanraj654 nutanraj654 · Answer 2 · 2022-07-16T12:30:56+02:00

a) a_2 =g-\frac{m_1}{m_2}(a_1 + \mu_k g)

Let's write the equation of the forces for the two blocks.

For block m1, the one which is on the horizontal table, there are two forces: the tension of the string, T, and the force of friction, \mu_k m_1 g, pulling in the opposite direction. So the equation of motion is

T-\mu_k m_1 g = m_1 a_1 (1)

For block m2, the one which is hanging vertically, there are two forces: the weight m_2 g, pushing downward, and the tension in the cable, acting upward. So the equation of the forces is

m_2 g - T = m_2 a_2 (2)

(note that if the cable is inextensible, the two blocks have the same acceleration, so a_1 = a_2).

We can isolate the tension T from eq.(1):

T=m_1 (a_1 +\mu_k g)

And substitute it into eq. (2):

m_2 g - m_1 (a_1 +\mu_k g) = m_2 a_2\\a_2 =g-\frac{m_1}{m_2}(a_1 + \mu_k g)

b) T=\frac{m_1 m_2g (1+\mu_k)}{m_1 + m_2}

Since the accelerations of the two blocks are equal: a_1 = a_2=a, we can rewrite the two equations of motion as

T=m_1 (a +\mu_k g) (1)

a=g-\frac{T}{m_2} (2)

So we can substitute the expression for a from (2) into (1) and solve the equation for T, the tension:

T=m_1 (g-\frac{T}{m_2})+\mu_kg)\\m_2 T=m_1 m_2 g - m_1 T + m_1 m_2 \mu_k g\\T=\frac{m_1 m_2g (1+\mu_k)}{m_1 + m_2}

c) 77.0 N

In order to find the magnitude of the tension, we just need to substitute data into the equation find at point b):

T=\frac{(9.4 kg)(11.5 kg)(9.8 m/s^2) (1+0.52)}{9.4 kg + 11.5 kg}=77.0 N

How do you find the magnitude of the acceleration of the system?

From the definition, acceleration is a rate of velocity change. If the initial velocity is v0 and the final velocity is v1, the acceleration arises as to the difference of these vectors divided by time interval Δt: a = (v1 - v0) / Δt.

Learn more about the magnitude of acceleration at

https://brainly.com/question/1597065

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