**Problem 5:**The massless spring of a spring gun has a force constant \( k = 1200 \, \text{N/m} \). When the gun is aimed vertically, a \( 0.015 \, \text{kg} \) projectile is shot to a height of \( 5.0 \, \text{m} \) above the end of the expanded spring (see figure below). How much was the spring compressed initially? Answer: \( 0.035 \, \text{m} \).**Diagram Explanation:**The diagram consists of three stages:1. **Initial Stage**: The spring is compressed by an unknown distance \( d \) and supports a projectile.2. **Intermediate Stage**: Upon release, the spring expands, exerting force on the projectile, projecting it upward.3. **Final Stage**: The projectile reaches a maximum height of \( 5.0 \, \text{m} \) from the top of the expanded spring.Each stage illustrates the projectile's movement and the state of the spring during compression, at release, and maximum projectile height. The image contains equations related to energy in a physical system. Here’s the transcription:- \( KE = \frac{1}{2}mv^2 \)- \( U_g = mgh \)- \( U_s = \frac{1}{2}kx^2 \)- \( E = KE + U_g + U_s \)- \( E_i = E_f \)- \( W_{fric} = F_{fric} d \)- \( F_{fric} = \mu_k N \)- \( E_i - W_{fric} = E_f \)Explanation:- \( KE \) is the kinetic energy, calculated as \( \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.- \( U_g \) is gravitational potential energy, given by \( mgh \), where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height.- \( U_s \) is elastic potential energy, expressed as \( \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement.- \( E \) represents total mechanical energy, the sum of kinetic and potential energies (\( KE + U_g + U_s \)).- \( E_i = E_f \) indicates initial energy equals final energy in the absence of non-conservative forces.- \( W_{fric} \) is the work done by friction, calculated as the product of the frictional force, \( F_{fric} \), and the distance, \( d \).- \( F_{fric} = \mu_k N \) describes the frictional force, where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.- \( E_i - W_{fric} = E_f \) accounts for energy loss due to friction, where initial energy minus work done by friction equals final energy.This set of equations is typically used in the study of energy conservation and the effects of friction in physics.

Answered: Image /qna-images/answer/53dbc68d-4175-4a88-9f77-3d8051d6c80a.jpg

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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