A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless pulley, as shown in figure (a). The coefficient of kinetic friction between the block and the surface is 0.300.

(a) Find the acceleration of the two objects and the tension in the string.
(b) Check the answer for the acceleration by using the system approach. (Use the following as necessary: m1, m2, μk, and g.)
(c) What if an additional mass is attached to the ball? How large must this mass be to increase the downward acceleration by 60%?

For (a) I calculated acceleration to be 5.10 m/s^2 and tension to be 36.2N.
For (b) I the equation is a = (m2g-ukm1g)/(m1+m2)
It is (c) that I cannot figure out. Can anybody please help me?

[tex]1.6a = \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3} \\ \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\ \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\ \\ m_3 = \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\ \\ m_3 = 22.57 kg[/tex]

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oeerivona oeerivona · Answer 2 · 2021-09-30T19:08:11+02:00

The force acting on objects connected by a string results in a tension in the string

(a) The acceleration of the two objects is approximately 5.106 m/s²

The tension in the string is approximately 36.2208 N

(b) [tex]The \ acceleration, \ a= \dfrac{Tension \ in \ string, T - Friction \ force,\, F_f}{Mass \ of \ block, \ m_1}[/tex]

(c) To increase the downward acceleration by 60%, the required additional mass to add is approximately 22.78 kg

The reasons why the above values are correct are as follows:

The given parameters are;

Mass of the block, m₁ = 4.50 kg

Mass of the ball, m₂ = 7.70 kg

Path over which the string connecting the two blocks passes = Frictionless pulley

Coefficient of friction between the block and the surface, μ = 0.300

Required:

(a) To find the acceleration of the two objects and the tension in the string

Solution:

The motion of the system is due to the weight of the ball, m₂·g

The opposing force to motion of the system is due to friction of the block, m₁·g·μ

The block and the ball connected by the string have a common acceleration given as follows;

(m₁ + m₂) × a = m₂·g - m₁·g·μ

Where;

a = The acceleration of the two objects

Therefore, we get;

[tex]a = \dfrac{g\cdot (m_2 - m_1\cdot \mu_k)}{m_1 + m_2}[/tex]

Which gives;

[tex]a = \dfrac{9.81 \times (7.70 - 4.50\times 0.300)}{4.50 + 7.70} \approx 5.106 \ m/s^2[/tex]

The acceleration of the two objects, a ≈ 5.106 m/s²

Tension in the string:

Tension in the string, T = m₂×(g - a)

∴ T = 7.70 × (9.81 - 5.106) ≈ 36.2208

The tension in the string, T ≈ 36.2208 N

(b) The force due tension in the string is given as follows

The force due to friction = 4.50 × 9.81 × 0.300 = 13.2435

Force due to friction, [tex]F_f[/tex] = 13.2435 N

Net force acting on block, F = T - [tex]F_f[/tex]

∴ F = 36.2208 N - 13.2435 N = 22.9773 N

The acceleration of the block, a, is given as follows;

[tex]a = \dfrac{22.9773}{4.5} \approx 5.106[/tex]

Therefore, the acceleration, a is correct as a ≈ 5.106 m/s²

(c) From the equation for acceleration, we have;

[tex]a = \dfrac{g\cdot (m_2 - m_1\cdot \mu_k)}{m_1 + m_2}[/tex]

When the acceleration is increased by 60%, we get

60% increase of a = 1.6·a, and the new mass of the ball = m₃, which gives;

[tex]1.6 \cdot a = \dfrac{g\cdot (m_3 - m_1\cdot \mu_k)}{m_1 + m_3}[/tex]

Plugging in the known values of a, m₁, g, and [tex]\mu_k[/tex] gives;

[tex]1.6 \times 5.106 = \dfrac{9.81\times (m_3 - 4.5\times 0.300)}{4.50 + m_3}[/tex]

1.6 × 5.106 × (4.50 + m₃) = 9.81 × (m₃ - 4.5×0.300)

8.1696·m₃ + 36.7632 = 9.81·m₃ - 13.2435

9.81·m₃ - 8.1696·m₃ = 36.7632 + 13.2435 = 50.0067

1.6404·m₃ = 50.0067

[tex]m_3 = \dfrac{50.0067}{1.6404} \approx 30.48[/tex]

The added mass Δm = m₃ - m₂

∴ Δm = 30.48 kg - 7.70 kg = 22.78 kg

The required additional mass to increase the downward acceleration by 60%, Δm = 22.78 kg

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