en Wiley.com/edugen/studenumi Halliday, Fundamentals of Physics, 10e Help l tem
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en Wiley.com/edugen/studenumi Halliday, Fundamentals of Physics, 10e Help l tem Announcements unread) ment Chapter 08, Problem 033 lower end of the incline is distance D 1.0 m from the In the figure, a spring with spring constant 160 Nm is at the top of a frictionless incline of angle e- 31e. The 0.24 m and released from rest, (a) What is end of the spring, which is at its relaxed length. A2.0 kg pushed against the spring until the spring is compressed the spring)? Cb) What is the speed of the the speed of the at the instant the spring returns to its relaxed length (which is when the canister los contact with when reaches the lower of the incline? (a) Number Units (b) Number click if you would like to show work for this question open Showwerk SHOW HINT co TUTORIAL LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEo MINI-LECTURE LINK TO TEXT 4.22.1 2olicy I 2000-20 lohn Wiley s Sons Inc. All Rights Reserved A Division of Rohn Wiley s sons Inc. nment/testlaglisteuniaN1 0064Explanation / Answer
This is place to use the principle of conservation of energy. The canister is (1.0+0.24) m=1.24 m from the bottom before being released. That gives the canister gravitational potential energy given by
GPE = m*g*h
where m = 2.0 kg, g = 9.8 m/s^2, and h = cos 31° x 1.24 m. The combination of units that this will give you is equivalent to Joules. The compressed spring gives additional potential energy given by
SPE = (1/2)*k*x^2
where k = 160 N/m, and x = 0.24 m. This result is also in Joules.
The total energy before release will be converted 100% to kinetic energy after the canister reaches the bottom. The kinetic energy is given by
KE = (1/2)*m*v^2
or, 2.0x9.8xcos31 x1.24 + 0.5x160x0.24^2 = 0.5x2.0xv^2
or, v = 5.04 m/s .........................................................ans
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