51 •51 GO Figure 5-47 shows two blocks connected bya cord (of negligible mass) that passes over a fric-tionless pulley (also of negligible mass). Thearrangement is known as Atwood's machine. Oneblock has mass m¡ = 1.30 kg; the other has mass m22.80 kg. What are (a) the magnitude of the blocks’ ac-celeration and (b) the tension in the cord?m1M2

Given: mass, m1 = 1.30 kg mass, m2 = 2.80 kg Solution:…

Answered: •51 GO Figure 5-47 shows two blocks…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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•51 GO Figure 5-47 shows two blocks connected by
a cord (of negligible mass) that passes over a fric-
tionless pulley (also of negligible mass). The
arrangement is known as Atwood's machine. One
block has mass m¡ = 1.30 kg; the other has mass m2
2.80 kg. What are (a) the magnitude of the blocks’ ac-
celeration and (b) the tension in the cord?
m1
M2

Expert Solution

Step 1

Given:

mass, m₁ = 1.30 kg

mass, m₂ = 2.80 kg

Solution:

From the free-body diagram of the system, we get:

$T - m_{1} g = m_{1} a - 1 m_{2} g - T = m_{2} a - 2$

From equation 1 and 2 , we get:

$T = m_{1} a + m_{1} g - 3 T = m_{2} g - m_{2} a - 4$

Equating equation 3 and 4 , we have:

$\begin{array}{rcl} m_{1} a + m_{1} g & = & m_{2} g - m_{2} a \\ m_{1} a + m_{2} a & = & m_{2} g - m_{1} g \\ a (m_{1} + m_{2}) & = & g (m_{2} - m_{1}) \\ a & = & g (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) \end{array}$

Substituting the given values in above equation, we get:

$\begin{array}{rcl} a & = & (9.8 m / s^{2}) (\frac{2.80 kg - 1.30 kg}{2.80 kg + 1.30 kg}) \\ = & (0.366 \times 9.8) m / s^{2} \\ = & 3.59 m / s^{2} \end{array}$

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