An object of mass m 1 = 9.80 kg is in equilibrium when connected to a light spri

Question

An object of mass m1 = 9.80 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in part [a] of the figure below. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.230 m (see part [b] of the figure below). The system is then released and both objects start moving to the right on the frictionless surface.

(a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see part [c] in the above figure) and moves to the right with speed v. Determine the value of v.
v = .561 m/s

(b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in part (d) of the above figure)?

I cant get part b, apparently its around 11 ish cm but i get 4.6 when I attempt. What I did was take one fourth of the period of the spring multiplied by the velocity and then subtracted the compression distance...but this gives me 4.6 which is pretty far off I guess. should be around 11 so if someone could help I would appreciate it

Explanation / Answer

We will apply conservation of energy.

When the spring is compressed, it will gain some potential energy.
When the spring regains its natural length, it will do work on masses to increase their kinetic energy.
They will lose contact with each other when the spring comes to its natural length because after getting natural length, the spring will pull the m1 block whereas there is no one to pull m2.
Also, their velocities will be equal.
kx2/2 = m1v2/2 + m2v2/2 ---------- (1)
100(0.230)2= 9.8 v2 + 7.0 v2
100 x 0.0529 = 16.8v2
5.29 = 16.8v2
0.3148 = v2
v = 0.561 m/s

2) The spring stretches as much as it was compressed. So, the m1 mass will be 0.230 m away from the dotted line.

then,
We have to calculate time taken for the spring to stretch fully from natural length.
This one is a bit tedious.
First, we going to take one-fourth SHM of the spring block system.
This will mean the time taken for the block to reach 0.230 m distance from dotted line.
T = 2??(m/k)
T / 4 = ?/2?(m/k) = (3.14 / 2 ) ?(9.8 / 100)
= 0.491 seconds

In this time, m2 will travel distance = 0.491 x 0.561 = 0.275 m

So,
D = 0.275 - 0.230 = 0.045 m

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