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Calculus Exercise - Atwood's Machine The figure below shows Atwood's Machine, in which two containers filled with water are connected by a cord (with negligible mass) passing over a pulley whose bearings are essentially "frictionless". At time t=0 s, container 1 has mass 1.40 kg and container 2 has mass 2.80 kg, but container 2 is leaking mass at the constant rate R = dm2/dt = -0.200 kg/s. Note that, since m2 is decreasing at a constant rate, m2(t) decreases linearly with time as m2(t)=mo - Rt, where mo is the initial mass of m2 (2.80 kg). (a) Use Newton's 2nd Law to show that the accelerations of m₁ and m2 at time t, a(t), are given by a(t) = [(m2(t) — m1)/(m2(t) + m1)] g. where m2(t) is the mass of container 2 at time t. (b) What are the accelerations of m1 and m2 at t=0 s and t= 5.00 s? (answer: a(t=0 s) = 3.27 m/s²; a(t = 5.00 s) = 1.23 m/s²) (c) At what rate, da/dt, is the acceleration of the containers changing at t=0s and 1=5.00 s? (answer: da/dt = -0.289 m/s³ at t= 0; da/dt = -0.498 m/s³ at t= 5.00 s) (d) At what time do the accelerations of me and m2 reverse direction? That is, at what time do mi and m2 start decelerating before their motions eventually reverse direction? (answer: t=7.00 s) (e) At what time is the acceleration of m¹ and m2 at its maximum value? (answer: t = 14.0 s) 201

Calculus Exercise - Atwood's Machine The figure below shows Atwood's Machine, in which two containers filled with water are connected by a cord (with negligible mass) passing over a pulley whose bearings are essentially "frictionless". At time t=0 s, container 1 has mass 1.40 kg and container 2 has mass 2.80 kg, but container 2 is leaking mass at the constant rate R = dm2/dt = -0.200 kg/s. Note that, since m2 is decreasing at a constant rate, m2(t) decreases linearly with time as m2(t)=mo - Rt, where mo is the initial mass of m2 (2.80 kg). (a) Use Newton's 2nd Law to show that the accelerations of m₁ and m2 at time t, a(t), are given by a(t) = [(m2(t) — m1)/(m2(t) + m1)] g. where m2(t) is the mass of container 2 at time t. (b) What are the accelerations of m1 and m2 at t=0 s and t= 5.00 s? (answer: a(t=0 s) = 3.27 m/s²; a(t = 5.00 s) = 1.23 m/s²) (c) At what rate, da/dt, is the acceleration of the containers changing at t=0s and 1=5.00 s? (answer: da/dt = -0.289 m/s³ at t= 0; da/dt = -0.498 m/s³ at t= 5.00 s) (d) At what time do the accelerations of me and m2 reverse direction? That is, at what time do mi and m2 start decelerating before their motions eventually reverse direction? (answer: t=7.00 s) (e) At what time is the acceleration of m¹ and m2 at its maximum value? (answer: t = 14.0 s) 201

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**Calculus Exercise – Atwood’s Machine** The figure below shows *Atwood’s Machine*, in which two containers filled with water are connected by a cord (with negligible mass) passing over a pulley whose bearings are essentially “frictionless.” At time \( t = 0 \), container 1 has mass 1.40 kg and container 2 has mass 2.80 kg, but container 2 is leaking mass at the constant rate \( R = dm_2/dt = -0.200 \, \text{kg/s} \). Note that, since \( m_2 \) is decreasing at a constant rate, \( m_2(t) \) decreases *linearly* with time as \( m_2(t) = m_0 - Rt \), where \( m_0 \) is the initial mass of \( m_2 \) (2.80 kg). (a) Use Newton’s 2nd Law to show that the accelerations of \( m_1 \) and \( m_2 \) at time \( t \), \( a(t) \), are given by \[ a(t) = \left[ \frac{(m_2(t) - m_1)}{(m_2(t) + m_1)} \right] g. \] where \( m_2(t) \) is the mass of container 2 at time \( t \). (b) What are the accelerations of \( m_1 \) and \( m_2 \) at \( t = 0 \, \text{s} \) and \( t = 5.00 \, \text{s} \)? (answer: \( a(t = 0) = 3.27 \, \text{m/s}^2; \, a(t = 5.00 \, \text{s}) = 1.23 \, \text{m/s}^2 \)) (c) At what rate, \( da/dt \), is the acceleration of the containers changing at \( t = 0 \, \text{s} \) and \( t = 5.00 \, \text{s} \)? (answer: \( da/dt = -0.289 \, \text{m/s}^3 \) at \( t = 0; \, da/dt = -0.498 \, \text{m/s}^
Transcribed Image Text:**Calculus Exercise – Atwood’s Machine** The figure below shows *Atwood’s Machine*, in which two containers filled with water are connected by a cord (with negligible mass) passing over a pulley whose bearings are essentially “frictionless.” At time \( t = 0 \), container 1 has mass 1.40 kg and container 2 has mass 2.80 kg, but container 2 is leaking mass at the constant rate \( R = dm_2/dt = -0.200 \, \text{kg/s} \). Note that, since \( m_2 \) is decreasing at a constant rate, \( m_2(t) \) decreases *linearly* with time as \( m_2(t) = m_0 - Rt \), where \( m_0 \) is the initial mass of \( m_2 \) (2.80 kg). (a) Use Newton’s 2nd Law to show that the accelerations of \( m_1 \) and \( m_2 \) at time \( t \), \( a(t) \), are given by \[ a(t) = \left[ \frac{(m_2(t) - m_1)}{(m_2(t) + m_1)} \right] g. \] where \( m_2(t) \) is the mass of container 2 at time \( t \). (b) What are the accelerations of \( m_1 \) and \( m_2 \) at \( t = 0 \, \text{s} \) and \( t = 5.00 \, \text{s} \)? (answer: \( a(t = 0) = 3.27 \, \text{m/s}^2; \, a(t = 5.00 \, \text{s}) = 1.23 \, \text{m/s}^2 \)) (c) At what rate, \( da/dt \), is the acceleration of the containers changing at \( t = 0 \, \text{s} \) and \( t = 5.00 \, \text{s} \)? (answer: \( da/dt = -0.289 \, \text{m/s}^3 \) at \( t = 0; \, da/dt = -0.498 \, \text{m/s}^
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