You simply need to know the formula for the potential energy stored in a spring to solve this problem. The formula is:

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

where k is the spring constant, and x is the distance from equilibrium

We can rearrange this to get:

$\displaystyle x = \sqrt{\frac{2U}{k}}$

Plugging in our values, we get:

$\displaystyle x = \sqrt{\frac{2\cdot20J}{15\frac{N}{m}}} = \sqrt{2.\overline{66}m^2} = 1.63m$

Since the block is motionless, we know that our forces will cancel out:

$\displaystyle F_{net}=0$

There are three forces in play: one from each spring, as well as the force of gravity. If we assume that forces pointing up are positive, we can write:

$\displaystyle F_{spring,top}+F_{spring,bot} - mg = 0$

Plugging in expressions for each spring force, we get:

$\displaystyle kx_{top}+kx_{bot} - mg = 0$

Rearring for the displacement of the top spring, we get:

$\displaystyle x_{top}=\frac{mg-kx_{bot}}{k} = \frac{(3kg)(10\frac{m}{s^2})-(25\frac{N}{m})(0.4m)}{25\frac{N}{m}}$

$\displaystyle x_{top} = \frac{30N - 10N}{25\frac{N}{m}}= 0.8m$

Since the block is motionless, we can assume that the forces cancel out:

$\displaystyle F_{net}=0$

If we designate any forces pointing downward as positive, we can write:

$\displaystyle F_g +F_{spring,bot}-F_{spring,top}=0$

Inserting expressions for each force, we get:

$\displaystyle mg + kx_{bot}- kx_{top} = 0$

Rearranging for mass, we get:

$\displaystyle m = \frac{k(x_{top}-x_{bot})}{g} = \frac{10\frac{N}{m}(2.5m-0.75m)}{10\frac{m}{s^2}}$

$\displaystyle m = 1.75kg$

There are two ways to solve this problem: the first involves calculating the spring constant and the second does not. We'll go through both methods.

Calculating Spring Constant

We can use the expression for the force of a spring:

$\displaystyle F=kx$

At equilibrium, the force of the spring equals the force of gravity:

$\displaystyle mg = kx$

Rearranging for the spring constant and plugging in values, we get:

$\displaystyle k = \frac{mg}{x}=\frac{(2kg)(10\frac{m}{s^2})}{0.2m}= 100\frac{N}{m}$

Now, apply this equation when the spring is on a different planet:

$\displaystyle mg = kx$

Rearranging for displacement and plugging in values, we get:

$\displaystyle x = \frac{mg}{k}=\frac{(2kg)(4\frac{m}{s^2})}{100\frac{N}{m}} = 0.08m$

Without Calculating Spring Constant

We can write the force equation for each scenario. Let the subscript 1 denote Earth, and 2 denote the other planet:

$\displaystyle mg_1=kx_1$

$\displaystyle mg_2 = kx_2$

Using these equations, we can set up a proportion:

$\displaystyle \frac{mg_2}{mg_1} = \frac{kx_2}{kx_1}$

$\displaystyle \frac{g_2}{g_1}=\frac{x_2}{x_1}$

Rearranging for the displacement of scenario 2 and plugging in values:

$\displaystyle x_2 = x_1\left ( \frac{g_2}{g_1}\right ) = (0.2m)\left ( \frac{4\frac{m}{s^2}}{10\frac{m}{s^2}}\right )=0.08m$

Since the displacement of the spring is at equilibrium, we can write:

$\displaystyle F_{net}=0$

There are three forces we can account for: spring force, gravitational force, and the additional force resulting from the acceleration of the elevator. If we assume that forces pointing upward are positive, we can write:

$\displaystyle F_{spring} -F_g - F_e = 0$

If you are unsure whether the force resulting from the acceleration of the elevator will be positive or negative, think about the situtation from personal experience: When an elevator begins to accelerate upward, your body feels heavier. Thus, the force adds to the normal gravitational force.

Substituting expressions for each force, we get:

$\displaystyle kx - mg - ma = 0$

Rearrange to solve for displacement:

$\displaystyle x = \frac{m(g+a)}{k} = \frac{(5kg)\left (10\frac{m}{s^2}+3\frac{m}{s^2} \right )}{60\frac{N}{m}}$

$\displaystyle x = 1.08m$

The maxmimum force applied by the spring will occur when the mass is at its maximum displacement. Since we know the energy of the system, we can calculate displacement using the following expression:

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

Rearranging for displacement, we get:

$\displaystyle x = \sqrt{\frac{2U}{k}}$

We can use this, along with the expression for force applied by a spring:

$\displaystyle F = kx$

Substitute our first expression into our second and simplify:

$\displaystyle F = k\sqrt{\frac{2U}{k}} = \sqrt{2Uk}$

We have values for each variable, allowing us to solve for the force:

$\displaystyle F = \sqrt{2(10J)(20\frac{N}{m})} = \sqrt{400N^2}=20N$

Note that the mass of the block is irrelevant to the problem. Mass does not effect displacement or force applied by the spring; it only affects the velocity of the block at different points.

We simply need to alter the expression for the force of the springs to solve this problem. The following is the original expression:

$\displaystyle F = kx$

Since each spring has the same costant, they actually act as one large spring with the same, original costant. Therefore the value of $\displaystyle x$ in this equation is the total displacement. Multiplying the displacement of each single spring by the number of total springs will give us this total displacement:

$\displaystyle F = k(x_sn)$

Here, $\displaystyle {n}$ is the number of springs, and this new displacement, $\displaystyle {x_s}$, is the displacement of each spring. Rearranging for the number of springs, we get:

$\displaystyle n = \frac{F}{kx_s} = \frac{20N}{(0.5m)(5\frac{N}{m})} = 8\ springs$

We can use force equilibrium to begin our derivation:

$\displaystyle F_{net}=0$

There are two vertical forces in play: gravity and total spring force. Since net force is zero, we know that these two general forces are equal to each other:

$\displaystyle F_{spring}=F_g$

Substituting in expressions for each force, we get:

$\displaystyle 40(kxsin(\theta)) = mg$

There are two things to note about the total spring force. The first is that we multiply it by 40 because there are 40 individual springs. Second, we multiply the force by the sine of the angle, because we only want to know the vertical force applied by the springs.

Rearranging for the displacement of the springs, we get:

$\displaystyle x = \frac{mg}{40ksin(\theta)}$

We have values for each variable, allowing us to solve:

$\displaystyle x = \frac{(50kg)(10\frac{m}{s^2})}{40(100\frac{N}{m})sin(20^{\circ})}$

$\displaystyle x = 0.37m$

The base equation for position when undergoing simple harmonic motion is:

$\displaystyle x(t) = Asin(\omega t+\phi)$

$\displaystyle \omega = \sqrt{\frac{k}{m}}$

First, solve for the phase constant.

$\displaystyle x(0) = Asin(\phi) = 5, \phi = \frac{\pi}{2}$

Plug all the variables into the equation and solve.

$\displaystyle x(3) = 5sin(\sqrt{\frac{35}{5}}3+\frac{\pi}{2}) = -0.42 cm$

Recall Hooke's Law, which states:

$\displaystyle F=-kX$

Here, $\displaystyle F$ is the force exerted by the spring, $\displaystyle k$ is the spring constant, and $\displaystyle X$ is the displacement from the spring's rest position. This equation tells us that the force exerted is directly proportional to the displacement. We don't need to solve for $\displaystyle k$ to determine the magnitude of the force on the spring stretched 10m. We can instead come up with a proportionality such that:

$\displaystyle \frac{F_1}{X_1}=\frac{F_2}{X_2}$

Here, $\displaystyle F_1$ and $\displaystyle F_2$ are forces applied on the string and $\displaystyle X_1$ and $\displaystyle X_2$ are the displacements of the spring from its respect position respectively. We assume that a stretched spring will have a positive displacement, whereas a shrunken spring will have a negative displacement. However, since we're looking for the magnitude of the force, regardless of direction, the direction of the displacement doesn't matter. Therefore, we can write the proportion as:

$\displaystyle \mid \frac{F_1}{ X_1}\mid=\mid \frac{F_2}{ X_2 }\mid$

In our case:

$\displaystyle \mid \frac{F_{unknown}}{10m}\mid =\mid \frac{20N}{-5m} \mid$

$\displaystyle \frac{F_{unknown}}{10m}=\frac{20N}{5m}$

$\displaystyle F_{unknown}=40N$