workingkinda
Answered

An old truck compresses a spring scale at the junkyard by 25 cm. The spring constant for the industrial scale at the junkyard is 39,200 N/m. What is the force applied to the scale by the car? N

Answer :

cryssatemp

Answer: 9800 N

Explanation:

According to Hooke's law, the elongation of a spring is directly proportional to the modulus of the force [tex]F[/tex] applied to it, as long as the spring is not permanently deformed. This law is mathematically defined as follows:

 

[tex]F=k \Delta x[/tex]

Where:  

[tex]k=39200 N/m[/tex] is the elastic constant of the spring  in this situation

[tex]\Delta x=25 cm \frac{1m}{100 cm}=0.25 m[/tex] is the measure of the compression of the spring

Solving the equation:

[tex]F=(39200 N/m)(0.25 m)[/tex]

[tex]F=9800 N[/tex]  This is the force applied to the scale

shawnwelch267

Answer: 9800 N

Explanation:

According to Hooke's law, the elongation of a spring is directly proportional to the modulus of the force applied to it, as long as the spring is not permanently deformed. This law is mathematically defined as follows:

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