Go Back   Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Feb 22nd 2009, 02:53 PM   #1
Junior Member
 
Join Date: Jan 2009
Posts: 12
Question 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 ?
baytom is offline   Reply With Quote
Old 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.
tashirosgt is offline   Reply With Quote
Old 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
mircea is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Tags
constant, spring



Thread Tools
Display Modes


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


Facebook Twitter Google+ RSS Feed