- #1

Ebby

- 41

- 14

- Homework Statement
- What will be the maximum instantaneous deformation of the springs if the car is lifted by a crane and dropped on the street from a height of 0.8 m?

- Relevant Equations
- W = 0.5*kx^2

K.E. = 0.5*mv^2

G.P.E. = mgh

I approach this by considering the four springs in parallel each with spring constant ##k## as one spring with four times the spring constant ##k' = 4k##. The car is dropped and at the moment its tyres touch the ground I assume that the spring is in its resting position. As the car continues to travel in a downward direction the spring is loaded up, absorbing the car's kinetic and gravitational potential energy.

The governing equations seems to be:$$P.E._{spring} = K.E._{car}+\,G.P.E._{car}$$

$$k'y^2 = mv^2 + 2mgy$$

So we have a quadratic equation involving ##y##:$$k'y^2 - 2mgy - mv^2 = 0$$

Where: ##v^2 = 2gs = 2(9.81)( 0.8)##. So:$$280000y^2 - 2(1200)(9.81)y - 1200(2)(9.81)( 0.8) = 0$$

$$280000y^2 - 23544y - 18835.2 = 0$$

This has a positive solution at ##y = 0.31##. However, I know that the correct answer is ##0.26## (metres). I cannot see what I've done wrong. I am pretty sure I have got the governing equation right, but... Any ideas?