A block of mass \(m_{1}=3.0 \mathrm{kg}\) rests on a frictionless horizontal surface. A second block of mass \(m_{2}=2.0 \mathrm{kg}\) hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block. The blocks are released from rest. (a) Find the acceleration of the two blocks after they are released. (b) What is the velocity of the first block 1.2 s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor? (c) How far has block 1 moved during the 1.2 -s interval? (d) What is the displacement of the blocks from their initial positions 0.40 s after they are released?

Step by step solution

01

Identify the forces acting on the system

Consider both blocks as a whole system. For mass \(m_1\), there is a tension force \(T\) acting on it horizontally. For mass \(m_2\), there is a gravitational force \(m_2g\) acting downwards and tension force \(T\) acting upwards.

02

Apply Newton's second law of motion to each block

For mass \(m_1\), we have: \(m_1 a_1 = T\), (considering the horizontal direction as positive) For mass \(m_2\), we have: \(m_2 a_2 = m_2g - T\), (considering the vertical direction as positive)

03

Set up the equation for the total acceleration of the system

Since the cord is inextensible, the acceleration of both blocks is the same: \(a_1 = a_2 = a\). Now we can rewrite our equations as follows: For mass \(m_1\), \(m_1 a = T\), For mass \(m_2\), \(m_2 a = m_2g - T\).

04

Solve the acceleration

From the equations, we can find the acceleration: Combine both equations by eliminating \(T\): \(m_1a + T = m_2g - T \Rightarrow a(m_1 + m_2) = m_2 g\) Now solve for \(a\): \(a = \frac{m_2 g}{m_1 + m_2} = \frac{(2.0\,\text{kg})(9.81\,\text{m/s}^2)}{3.0\,\text{kg} + 2.0\,\text{kg}}\) \(a = 3.92 \, \text{m/s}^2\)

05

Calculate the velocity of the first block after 1.2s

Use the equation \(v = u + at\), assuming the initial velocity \(u\) is 0: \(v = 0 + (3.92\, \text{m/s}^2)(1.2\,s)\) \(v = 4.70 \,\text{m/s}\)

06

Calculate the displacement of block 1 after 1.2s

Use the equation \(s = ut + \frac{1}{2}at^2\), assuming the initial velocity \(u\) is 0: \(s = 0 + \frac{1}{2}(3.92\, \text{m/s}^2)(1.2\,s)^2\) \(s = 2.82\,\text{m}\)

07

Calculate the displacement of both blocks after 0.40s

For block 1: \(s_1 = 0 + \frac{1}{2}(3.92\, \text{m/s}^2)(0.40\,s)^2\) \(s_1 = 0.314\,\text{m}\) For block 2: \(s_2 = 0 - \frac{1}{2}(3.92\, \text{m/s}^2)(0.40\,s)^2\) \(s_2 = -0.314\,\text{m}\) The negative sign indicates that block 2 moves downwards.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!