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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 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) + mi)] g. where m2(t) is the mass of container 2 at time t. (b) What are the accelerations of mı and m2 at t= 0s 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=0 s and t= 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 m₁ 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) 181 M₂
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 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) + mi)] g. where m2(t) is the mass of container 2 at time t. (b) What are the accelerations of mı and m2 at t= 0s 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=0 s and t= 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 m₁ 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) 181 M₂
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