Spring to the top

This diagram is extremely confusing---I thought it meant that the mass given was simply 2m high (like the previous problem, I thought it was extraneous information). Apart from this, the problem is relatively simple.

By conservation of energy: $\sum_{T_0} T_0+\sum_{V_0} V_0=\sum_{T_f} T_f+\sum_{V_f} V_f$ Clearly, then, kinetic energy: $T_0=0\text{ J}$ And potential energy: $V_0=\frac{1}{2}kl^2$ Where we have: $l=x_f-x_0=10\sqrt{2}-8$ Then: $V_0=\frac{1}{2}k(10\sqrt{2}-8)^2$ At its normal position: $V_f=\frac{1}{2}k(2^2)$ This give us: $T_f=V_0-V_f=\frac{1}{2}(100)(10\sqrt{2}-8)^2-\frac{1}{2}(100)(2^2)\approx 1686.2\text{ J}$ Which is our answer.

By the Work-Energy Theorem, $W = \Delta K$ . Since the initial kinetic energy is 0, the final kinetic energy is simply the change in the kinetic energy, or the work done by the mass on the spring. Since the force within the spring is variable, we must use integration to compute the work: $W = \int F \cdot {dx}$ . The magnitude of the force exerted by mass on the spring is $F = k(l - 8)$ , where l is the length of the spring (8 is the natural length of the spring. However, since our given values are in terms of x (the horizontal distance from the beginning), we must make the substitution $l = \sqrt{x^2+100}$ . We get $F(x) = k(\sqrt{x^2+100}-8)$ Now we take the integral $\int_{x=0}^{10} F(x) \cdot {dx}$ . Recall that when we take the dot product of two vectors, we multiply their magnitudes together with the cosine of the angle between them. The cosine of the angle between the force in the spring and the differential of the horizontal displacement is $\dfrac{x}{\sqrt{x^2+100}}$ , therefore, we can rewrite our integral as $W = 100 \int_{x=0}^{10} \dfrac{x(\sqrt{x^2+100} - 8){dx}}{\sqrt{x^2+100}}$ (the spring constant $k=100$ , is put in front of the integral for simplicity) This integral evaluates to 1686.2 J, the correct answer.

When the mass is pulled, it has elastic potential energy. When it returns to its starting point, it has elastic potential energy and translational kinetic energy.

By principle of conservation of energy, $\sum{E_{initial}}=\sum{E_{final}}$ .

$\frac{1}{2}kx_i^2=\frac{1}{2}kx_f^2+E_k$

$\frac{1}{2}(100)(\sqrt{200}-8)=\frac{1}{2}(100)(10-8)+E_k$

$E_k=1686.2J$

By conservation of mechanical energy, $KE_1 + PE_1 = KE_2 + PE_2$ . $KE_1=0$ , so $KE_2 = PE_1-PE_2$ . Doing some trigonometry, we find that the length of the extended spring is $10\sqrt 2$ , so the potential energy there is $\frac{1}{2} (100)(10\sqrt 2-8)^2$ . Thus the kinetic energy is $\frac{1}{2}(100)(10\sqrt 2 - 8)^2 - \frac{1}{2}(100)(2)^2 = 1686.214$ J.

When the spring is pulled over a horizontal distance of 10m, its length is $\sqrt{10^{2}+10^{2}}$ = $\sqrt{200}$

Its extended length is x= $\sqrt{200}$ -8 . Its total potential energy is $\frac{1}{2}$ k $x^{2}$ =1886.29J

When return to starting point, the spring still has extension of 10-8=2 m and potential energy= $\frac{1}{2}$ k( $2^{2}$ )=200J . So when the spring returns to starting point, the portion of potential energy converted to kinetic energy is 1886.29-200= $\boxed{1686.29J}$

(KE+PE)initial = (KE+PE)final by Law of Conservation of Energy 0+ \frac{1}{2}ke {1}^2 = KE + \frac{1}{2}ke {2}^2 \frac{1}{2}(100)(10\sqrt{2} - 8)^2 = KE+ \frac{1}{2}(100)(10-8)^2 KE = 50(2^2)- 50(10\sqrt{2} - 8)^2 = 1686.2 J