Question

What should be the spring constant k of a spring designed to bring a 1300 kg car to rest from a speed of 95 km/h so that the occupants undergo a maximum acceleration of 5.0g?

Answer 1

Answer:

Spring constant, k = 4485.24 N/m

Explanation:

It is given that,

Mass of the car, m = 1300 kg

Speed of the car, v = 95 km/h = 26.38 m/s

Acceleration, a = 5 g

We need to find the spring constant k of the spring such that,

m a = k x

$x=\dfrac{ma}{k}$.........(1)

The elastic potential energy is balanced by the kinetic energy of the spring such that,

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$mv^2=\dfrac{(ma)^2}{k}$

$k=\dfrac{ma^2}{v^2}$

$k=\dfrac{1300\ kg\times (5\times 9.8)^2}{(26.38\ m/s)^2}$      

k = 4485.24 N/m

So, the spring constant of the spring is 4485.24 N/m. Hence, this is the required solution.  

Answer 2

We have that for the Question ,it can be said that he spring constant k of a spring designed to bring a 1300 kg car to rest from a speed of 95 km/h so that the occupants undergo a maximum acceleration of 5.0g is

k=4449.3Nm^{-1}

From the question we are told

What should be the spring constant k of a spring designed to bring a 1300 kg car to rest from a speed of 95 km/h so that the occupants undergo a maximum acceleration of 5.0g?

Generally the equation for the  angular frequency is mathematically given as

$\omega=\frac{a}{v}\\\\\omega=\frac{5*9.8}{95*\frac{1000}{3600}}\\\\\omega=1.85$

Generally the equation for the  spring constant k is mathematically given as

$k=\omega^2*m\\\\k=(1.85)^2*1300$

k=4449.3Nm^{-1}

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