Physics Help Forum Spring Constant

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Feb 22nd 2009, 02:53 PM #1 Junior Member   Join Date: Jan 2009 Posts: 12 Spring Constant Could someone please show me how to do this? What should be the spring constant of a spring designed to bring a 1500 car to rest from a speed of 95 so that the occupants undergo a maximum acceleration of 4.5 ?
 Feb 22nd 2009, 10:16 PM #2 Senior Member   Join Date: Dec 2008 Location: Las Cruces NM Posts: 256 Shall we imagine a compressed spring that uncoils or a stretched string that contracts? I'll imagine a stretched spring that is stretched to length L beyon it's natural length and has spring constant k. Let x = 0 be the initial position of the moving end of the spring at time t= 0. By Hook's law the force the spring exerts is $\displaystyle F_s = k (L - x)$ for $\displaystyle 0 \le x \le L$. The max force occurs when $\displaystyle x = 0$. Then $\displaystyle F_{max} = k L$. We have a limit on acceleration. $\displaystyle a_{max} = 4.5 g$. Newton says that $\displaystyle F_{max} = m a_{max}$ where $\displaystyle m$ is the mass of the car.. So must make $\displaystyle k L \le m a_{max}$ The work done by the spring when it snaps back to its natural length will be the kinetic energy gained by the car. Work is the integral of force as a function of distance. $\displaystyle \int_0^L k(L-x) dx = \frac{1}{2} m v^2$ where $\displaystyle v$ is the final velocity of the car. $\displaystyle \Big |_0^L (kLx - \frac {kx^2}{2}) = \frac{1}{2} m v^2$ $\displaystyle ( kLL - \frac{kL^2}{2}) - ( kL(0) - \frac{(k(0^2)}{2}) = \frac{1}{2} m v^2$ $\displaystyle \frac {kL^2}{2} = \frac{1}{2} m v^2$ ILet's make the spring constant large enough so that it produces the max acceleration. Then we have two equations in the unknowns $\displaystyle k$ and $\displaystyle L$: eq 1. $\displaystyle \frac {kL^2}{2} = \frac{1}{2} m v^2$ and eq 2. $\displaystyle k L = m a_{max}$ Solve for $\displaystyle k$ in eq. 2 and put the result in eq 1. $\displaystyle k = \frac {m a_{max}}{L}$ $\displaystyle \frac {m a_{max}}{L} \frac{L^2}{2} = \frac{1}{2} m v^2$ $\displaystyle m a_{max} L = m v^2$ $\displaystyle L = \frac{v^2}{a_{max}}$ You'll have to check my work.
 Feb 23rd 2009, 10:48 AM #3 Member   Join Date: Dec 2008 Posts: 65 I think the question was "what is the spring constant?". Conservation of energy gives: 1/2*m*v^2=1/2*k*xm^2 where xm is the maximum spring deformation (x=lo-l_final) The maximum acceleration happens when the force is maximum: m*a_max=k*xm Combining the two equations, k=m*a_max^2/v^2

 Tags constant, spring

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Nousher Periodic and Circular Motion 3 Feb 9th 2015 04:48 AM maserati Advanced Mechanics 2 Sep 19th 2013 06:17 PM tag16 Energy and Work 4 Jun 21st 2009 01:36 PM vincisonfire Energy and Work 2 Nov 22nd 2008 02:52 PM chocolatelover Kinematics and Dynamics 0 Nov 17th 2008 08:14 PM