44.2k views
0 votes
By what amount has the spring has been stretched when the acceleration of the boxes has magnitude 1 m/s2 ? assume gravitational acceleration is 10 m/s2 . answer in units of m.

User ALM
by
7.1k points

1 Answer

4 votes

Final answer:

Using Hooke's Law, the spring constant is first calculated with the formula k = F / x, and then used to find the spring stretch for an acceleration of 1 m/s² which results in a stretch of 0.008 m or 8 mm.

Step-by-step explanation:

To determine how much the spring has been stretched when the acceleration of the boxes is 1 m/s2, we first need to use the force constant (k) of the spring provided in one of the reference scenarios. According to the given information, the spring stretches 8.00 cm when a 10.0 kg load is applied. Since the force due to gravity (Fg) on this load is equal to the weight of the load (mass times gravitational acceleration), we can calculate Fg = 10.0 kg × 10 m/s2 = 100 N. Knowing this force and the stretch (x), we can find the force constant (k) of the spring using Hooke's law, which states that F = kx.

So we rearrange Hooke's law to solve for k: k = F / x = 100 N / 0.08 m = 1250 N/m. To find the stretch of the spring due to an acceleration of 1 m/s2, we again use F = ma, where m is the mass and a is the acceleration. The force causing this acceleration is F = 10.0 kg × 1 m/s2 = 10 N. Finally, to find the new stretch (x), we rearrange Hooke's Law to x = F / k = 10 N / 1250 N/m, which gives x = 0.008 m or 8 mm.

User Nik
by
6.6k points