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A soapbox derby race car starts at rest at the top of a track that has a vertical drop of 15 m. The car is to be stopped at the end of the track by colliding with and compressing a spring. In order to avoid injury to the driver, the spring must be compressed to 3.0 m before the car stops. What should be the force constant of this spring

Answer :

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

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