Answer:
V= 1.82 m/s
Explanation:
Given that
mass , m = 2840 kg
Spring constant ,K = 6 x 10⁶ N/m
Compression in the spring , x = 3.98 cm
Lets take speed of the car before impact = V
Now by using energy conservation
[tex]\dfrac{1}{2}Kx^2=\dfrac{1}{2}mV^2[/tex]
[tex]Kx^2=mv^2[/tex]
[tex]V=\sqrt{\dfrac{Kx^2}{m}}[/tex]
Now by putting the values in the above equation
[tex]V=\sqrt{\dfrac{6\times 10^6\times (3.98\times 10^{-2})^2}{2840}}[/tex]
V= 1.82 m/s
Therefore the speed of the car before impact will be 1.82 m/s.