Question

A spring is stretched a distance of Ax 40 cm beyond its relaxed length of x by a force, F-25 N. Attached to the end of the spring is an block of mass m-9 kg, which rests on a horizontal frictionless surface. After a moment, the force is removed. 1) What is the potential energy of the block due to the spring when it is held in the stretched position? Uclastic Submit 2) What is the total mechanical energy of the block when it is held in the stretched position? ME,- Submit 3) What is the total mechanical energy of the block when it returns to its unstretched position? ME2 Submit 4) When the spring again returns to its unstretched length, what is the speed of the block? Vo m/s Submit Help 5) What is the total mechanical energy of the block when it has returned only halfway (20 cm) to its unstretched position? ME Submit 6) Vhen the spring has returned only halfway (20 cm), what is the speed of the block? vu

Answer 1

First, we calculate the elastic constant of the spring. Remember that:

$F = kx \rightarrow k = \frac{F}{x} = \frac{25}{0.400} = 62.5 N/m$

1) The potential energy can be calculated as:

$U = \frac{1}{2}kx^2$

$U = \frac{1}{2}(62.5)(0.400)^2 = 5.0 J$

2) at his position v = 0, then K = 0

3) Due the surface is frictionless, there is no lost of energy, so the total mechanic energy is conserved, then:

4) In this case: , so , then:

$ME_2 = K_2 +U_2 = K_2 +0 = \frac{1}{2}mv^2_0$

solving for vo:

$v_0 = \sqrt{\frac{2ME_2}{m}}$

$v_0 = \sqrt{\frac{2(5)}{9}} = 1.05 m/s$

5) Due the conservation of energy, the total mechanics energy at this position:

6) In this position, x = 0.200 m, so the potential energy is giving by:

$U_3 = \frac{1}{2}kx^2 = \frac{1}{2}62.5(0.200)^2 = 1.25 J$

$K_3 = \frac{1}{2}mv^2$

then:

$1.25 +\frac{1}{2}(9)v^2 = 5.0 J$