Question

When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s

springs are compressed by 3.0 cm.

1.2.a. What is the spring constant of the car’s springs assuming they act as a single spring?
1.2.b. How far will the car lower if loaded with 3.00 x 102 kg rather than 2.00 x 102 kg​

Answer 1

Answer:

a

 $k = 457333.3 N/m$

b

$x_a =0.09\ m$      

Explanation:

From the question we are told that

    The total mass of  three people is  $M = 2.00*10^{2} \ kg$

     The mass of the car is  $m_c = 1.200 *10^{3} \ kg$

     The compression of the car spring is  $x = 3 \ cm = 0.03 \ m$

     

Generally the spring constant is mathematically represented as

          $k = \frac{F}{x}$

Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as        

=>       $F = (M +m_c) *g$

=>       $F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8$

=>       $F = 13720 \ N$

So

        $k = \frac{13720}{0.03}$

=>     $k = 457333.3 N/m$

Generally if the mass which the car is loaded with is  $m = 3.00*10^{2} \ kg$

Then the force experienced by the spring is  

         =>       $F_a = (m +m_c) *g$

         =>       $F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8$

         =>       $F_a = 41160 \ N$

Generally from the above formula the compression is  

       $x_a = \frac{F_a}{k}$

=>    $x_a = \frac{41160}{457333.3}$

=>    $x_a =0.09\ m$