A 60 kg student is standing atop a spring in an elevator that is accelerating upward at 3.0 m/s2. The spring constant is 2.5 x 103 N/m. By how much is the spring compressed?

the spring will be compressed by 0.3072 m

Explanation:

acceleration of elevator=3 m/s²

mass of student= 60 Kg

spring constant=2.5 x 10³ N/m

the force on the student is given by F = m ( g +a)

F=60 (9.8+3)

F=768 N

now the formula for spring force is given by

F= k x

768= 2.5 x 10³ (x)

x=0.3072 m

Final answer:

The spring on which a 60kg student is standing in an elevator accelerating at 3.0 m/s² is compressed by approximately 30.72 cm. This is calculated using Hooke's Law, considering both the weight of the student and the additional force due to the elevator's acceleration.

Explanation:

The situation you've described involves Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. In this case, we can consider the student's weight plus the extra force from the acceleration of the elevator.

To determine the compression of the spring, we use the equation F = kx, where F is the total force exerted on the spring, k is the spring constant, and x is the amount the spring is compressed. Here, the total force (F) includes the weight of the student and the force due to the elevator's acceleration. So, F = mg + ma, where m is the mass of the student, g is the gravity (9.8 m/s²), and a is the acceleration of the elevator.

Substituting the given values, we get F = (60 kg)(9.8 m/s²) + (60 kg)(3.0 m/s²) = 768 N. The compression of the spring (x) is now obtained by rearranging the Hooke's law formula to x = F/k.

This results in x = 768 N ÷ 2.5 x 10³ N/m = 0.3072 m or 30.72 cm. Thus, the spring is compressed by approximately 30.72 cm due to the combined force of the student's weight and the accelerating elevator.

Learn more about Hooke's Law here:

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