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Sagot :
In order to avoid injury to driver, The force constant of this spring should be 73.82 N/m
Calculation of force constant to this spring:
Here
vertical drop is 15 m.
mass of the car, m= 113 kg
gravitational acceleration, g= -9.8 m/s^2
spring compression before car stops, x= 3.0 m
In order to avoid injury to the driver,
we need to use Hooke's law and Newton's second law:
F(net) = F(x)
m*a=-k*x
here a=g
k= -(m*g) / x
here k is force constant of this spring
substituting the values we get,
k = -(113* -9.8) / (15)
k = 73.8266 N/m
Hence, the force constant of this spring should be 73.82 N/m
learn more about force constant of this spring here:
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