ballar9bod

2023-03-30

The spring in the figure (a) is compressed by length delta x . It launches the block across a frictionless surface with speed v0. The two springs in the figure (b) are identical to the spring of the figure (a). They are compressed by the same length delta x and used to launch the same block. What is the block's speed now?

dalematealreypq3a

To solve this problem, let's consider the principles of conservation of energy and Hooke's law.
In both figures (a) and (b), the potential energy stored in the compressed spring is converted into kinetic energy as the block is launched. We can equate the potential energy with the kinetic energy to find the block's speed.
The potential energy stored in the compressed spring is given by the formula:
$PE=\frac{1}{2}k\left(\Delta x{\right)}^{2}$
where $k$ is the spring constant and $\Delta x$ is the compression of the spring.
The kinetic energy of the block is given by:
$KE=\frac{1}{2}m{v}_{0}^{2}$
where $m$ is the mass of the block and ${v}_{0}$ is the initial velocity.
According to the conservation of energy, the potential energy is equal to the kinetic energy:
$\frac{1}{2}k\left(\Delta x{\right)}^{2}=\frac{1}{2}m{v}_{0}^{2}$
Now, let's move on to figure (b):
In this case, we have two identical springs compressed by the same length $\Delta x$. The potential energy stored in each spring is:
$P{E}_{1}=\frac{1}{2}k\left(\Delta x{\right)}^{2}$
$P{E}_{2}=\frac{1}{2}k\left(\Delta x{\right)}^{2}$
The total potential energy from both springs is:
$P{E}_{\text{total}}=P{E}_{1}+P{E}_{2}=\frac{1}{2}k\left(\Delta x{\right)}^{2}+\frac{1}{2}k\left(\Delta x{\right)}^{2}=k\left(\Delta x{\right)}^{2}$
Using the conservation of energy again, the total potential energy is equal to the kinetic energy of the block:
$k\left(\Delta x{\right)}^{2}=\frac{1}{2}m{v}_{\text{new}}^{2}$
Now, let's compare the speeds of the block in figure (a) and figure (b). We can divide the equation for figure (b) by the equation for figure (a) to find the ratio of the speeds:
$\frac{\frac{1}{2}m{v}_{\text{new}}^{2}}{\frac{1}{2}m{v}_{0}^{2}}=\frac{k\left(\Delta x{\right)}^{2}}{\frac{1}{2}k\left(\Delta x{\right)}^{2}}$
Simplifying the equation:
$\frac{{v}_{\text{new}}^{2}}{{v}_{0}^{2}}=\frac{k\left(\Delta x{\right)}^{2}}{\frac{1}{2}k\left(\Delta x{\right)}^{2}}$
$\frac{{v}_{\text{new}}^{2}}{{v}_{0}^{2}}=2$
Taking the square root of both sides:
$\frac{{v}_{\text{new}}}{{v}_{0}}=\sqrt{2}$
Therefore, the block's speed with two springs is $\sqrt{2}$ times the speed with one spring.

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