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You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1300 kg car moving at 0.62 m/s is to compress the spring no more than 0.072 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

Respuesta :

Answer:

[tex]F=3470.2778\ N[/tex]

Explanation:

Given:

mass of the car, [tex]m=1300\ kg[/tex]

parking speed of the car, [tex]u=0.62\ m.s^{-1}[/tex]

compression of spring bumpers on the walls, [tex]\delta x=0.072\ m[/tex]

Using the equation of motion:

[tex]v^2=u^2+2.a.\delta x[/tex]

where:

[tex]v=[/tex] final speed of the car [tex]=0\ m.s^{-1}[/tex]

[tex]a=[/tex] acceleration of the car while compressing the spring (will be -ve since final velocity tends to zero)

[tex]0^2=0.62^2+2\times a\times 0.072[/tex]

[tex]a=-2.6694\ m.s^{-1}[/tex] (negative sign denotes that it is reducing the speed )

Now the force:

[tex]F=m.a[/tex]

[tex]F=1300\times 2.6694[/tex]

[tex]F=3470.2778\ N[/tex]