The massless spring of a spring gun has a force constant k= 1200 N/m. When the gun is aimed vertically, a 0.015 kg projectile is shot to a height of 5.0 m above the end of the expanded spring (see figure below). How much was the spring compressed initially? Answer: 0.035 m T d = ? NNNNNNNN 5.0 m
The massless spring of a spring gun has a force constant k= 1200 N/m. When the gun is aimed vertically, a 0.015 kg projectile is shot to a height of 5.0 m above the end of the expanded spring (see figure below). How much was the spring compressed initially? Answer: 0.035 m T d = ? NNNNNNNN 5.0 m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Problem 5:
The massless spring of a spring gun has a force constant k= 1200 N/m. When the gun is aimed
vertically, a 0.015 kg projectile is shot to a height of 5.0 m above the end of the expanded
spring (see figure below). How much was the spring compressed initially? Answer: 0.035 m
T
이
NNNNNNNN
NNNNNNNN
5.0 m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F916b3e43-3e70-4159-b0c4-23957999f35d%2F53dbc68d-4175-4a88-9f77-3d8051d6c80a%2Ftgzoro_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 5:
The massless spring of a spring gun has a force constant k= 1200 N/m. When the gun is aimed
vertically, a 0.015 kg projectile is shot to a height of 5.0 m above the end of the expanded
spring (see figure below). How much was the spring compressed initially? Answer: 0.035 m
T
이
NNNNNNNN
NNNNNNNN
5.0 m
![KE = ½-mv², Ug = mgh, U₁ = ½kx², E = KE + Ug + Us, Ei = Eƒ,
Wfric=Ffricd,
Ffric = MkN, E₁ - Wfric = Ef](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F916b3e43-3e70-4159-b0c4-23957999f35d%2F53dbc68d-4175-4a88-9f77-3d8051d6c80a%2Fwie1q9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:KE = ½-mv², Ug = mgh, U₁ = ½kx², E = KE + Ug + Us, Ei = Eƒ,
Wfric=Ffricd,
Ffric = MkN, E₁ - Wfric = Ef
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