Question

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 1

Answer: 9800 N

Step-by-step 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:

Answer 2

Answer: 9800 N

Step-by-step explanation:

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

`F=k \Delta x`

Where:

`k=39200 N/m` is the elastic constant of the spring in this situation

`\Delta x=25 cm (1m)/(100 cm)=0.25 m` is the measure of the compression of the spring

Solving the equation:

`F=(39200 N/m)(0.25 m)`

`F=9800 N` This is the force applied to the scale