Physics 11 (Hooks law)

Question

Spring is of the length of 22cm. if it is loaded with 2kg it gets stretched to 38cm. Calculate its spring constant. Make sure you draw an FBD and write Fnet=ma to begin this solution.

Answer 1

It'll be a bit challenging to draw the force diagram by text, but I'll give it a shot:

Fs

^

| __

- |

- 33cm

- |

- _|_

O

|

v

Fg

The - represent the spring, and the O represents the object attached to the spring.

So the FBD shows that the object has the force of gravity Fg exerting downwards on it, and the force of the spring Fs exerting it upwards. Since object is remains still (it has net 0 acceleration), then this must mean that Fnet = ma = m*0 = 0. This gives us that Fnet = Fs - Fg = 0, and therefore Fs = Fg. Then, since we know that Fs = kΔx and Fg = mg, we have:

kΔx = mg

k = mg/Δx

We know that m = 2kg, g = 9.8m/s², and Δx = 38cm - 22cm = 16cm. We also need to convert from cm to m so that we have the solution in form of N/m; 16cm = 0.16m. Plugging the values in we have:

k = (2kg)(9.8m/s²)/(0.16m) = 122.5 kg/s² = 122.5 N/m.