Question

on earth, a spring with constant ( k = 100n/m ) is in its equilibrium position. a mass of 300 g is hung from it. how far does the spring stretch?
0.00294 m
0.0294 m
2.94 m
29.4 m
question 3
you obtain a spring of constant ( k = 550n/m ). you squeeze the spring as hard as you can with your hand and it compresses 3.2 cm. what is your grip strength (in newtons), ie what force can you squeeze with?

Explanation:

Step1: Convert mass to kg

Given mass \(m = 300g=0.3kg\).

Step2: Calculate force due to gravity

Using \(F = mg\) (where \(g = 9.8m/s^{2}\)), we have \(F=(0.3kg)\times(9.8m/s^{2}) = 2.94N\).

Step3: Use Hooke's Law \(F=kx\) to find \(x\)

Rearranging for \(x\), we get \(x=\frac{F}{k}\). Substituting \(F = 2.94N\) and \(k = 100N/m\), we have \(x=\frac{2.94N}{100N/m}=0.0294m\).

Step1: Convert compression distance to meters

Given \(x = 3.2cm = 0.032m\).

Step2: Use Hooke's Law \(F = kx\)

Substituting \(k = 550N/m\) and \(x=0.032m\), we get \(F=(550N/m)\times(0.032m)=17.6N\).

Answer:

B. \(0.0294m\)

For the second question: