A 1.27 kg object is held 1.30 m above a relaxed, massless vertical spring with a
ID: 1542067 • Letter: A
Question
A 1.27 kg object is held 1.30 m above a relaxed, massless vertical spring with a force constant of 312 N/m. The object is dropped onto the spring. (a) How far does the object compress the spring? (b) Repeat part (a), but now assume that a constant air-resistance force of 0.840 N acts on the object during its motion. (c) How far does the object compress the spring if the same experiment is performed on the moon, where g 1.63 m/s2 and air resistance is neglected? This electronic presentation to be used with SERWAY/JEWETT. Physics for Scientists and Engineers with Modem, Hybrid (with Enhanced WebAssign Homework and eBook LOE Printed Access Card for Multi Term Math and Science), 8E. From SERWAY JEWETT. Physics for Scientists and Engineers with Modern, Hybrid (with Enhanced WebAssign Homework and eBook LOE Printed Access Card for Multi Term Math and Science), 8E. 3 2012 Brooks/Cole, a part of Cengage Learning, Inc. All rights reserved. Reproduced by permission. Text/images may not be modified or reproduced in any way without prior written permission of the publisher, www.cengage.com/permissions Potential EnergyExplanation / Answer
(a) let x= compression in the spring
we have to use the conservation of energy
PEgi+ KEi=PEgf+PEs+ KEf
PEgi= initial gravitational potential energy=m g (1.3+x)
PEgf= initial gravitational potential energy=0
Ki= initial kinetic eenrgy=0, KEf= final kinetic energy=0
PEs= potential energy of spring= 1/2 kx2.
so mg (1.3+x)+0=0+1/2 kx2+ 0
1.27(9.8)(1.3+x)= 1/2 (312)x2.
16.18 + 12.45x=156x2.
156x2.-12.45x-16.18=0
we solve this quadratic equation by quadratic formula and we get, X=0.3644 m.
(B) we have to again use the conservation of energy
PEgi+ KEi + Wf=PEgf+PEs+ KEf
Wf= work done by air friction
mg (1.3+x)+0-F x=0+1/2 kx2+ 0
1.27(9.8)(1.3+x)-0.840(1.3+x)= 1/2 (312)x2.
16.18 + 12.45x-0.840x-1.092=156x2.
156x2.-11.61x-15.08=0
we solve this quadratic equation by quadratic formula and we get, X=0.3503 m.
(c) we have to again use the conservation of energy. on moon we will use the value of g=1.63 m/s^2
PEgi+ KEi =PEgf+PEs+ KEf
mg (1.3+x)+0=0+1/2 kx2+ 0
1.27(1.63)(1.3+x)= 1/2 (312)x2.
2.69 + 2.07x=156x2.
156x2.-2.07x-2.69=0
we solve this quadratic equation by quadratic formula and we get, X=0.138m.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.