AP Physics 1 - Spring Force

Two identical, massless, springs are placed in series. A mass of is hung from them. After all oscillations have stopped, the total length is . Calculate the spring constant of an individual spring.

$980\frac{N}{m}$

$560\frac{N}{m}$

$490\frac{N}{m}$

$1960\frac{N}{m}$

$1120\frac{N}{m}$

Explanation

Each spring will be subject to the same force, and since they have the same spring constant, stretch the same amount. Thus:

Total stretch:

Stretch of one spring:

$\frac{10cm}{2}=5cm$

Use Hooke's law:

The force will be equal to the force of gravity on the mass:

Solve for :

$k=\frac{mg}{x}$

$k=\frac{5*9.8}{.05}$

$k=980\frac{N}{m}$

Two springs with unknown constants are hung from a ceiling. However, we do know that the constant of one is four times that of the other. A block of mass is attached to both springs and the system drops . If you then manually stretch the system another , what is the total potential energy stored in the weaker spring?

$g = 10\frac{m}{s^2}$

Explanation

Since the springs are added in parallel, we can simply add their constants together to get an equivalent constant.

$k_{eq}=k_1+k_2$

From the statement, we know:

Substituting this in, we get:

$k_{eq}=5k_1$

Now we can use Hooke's Law along with the information of how far the block causes the springs to stretch:

$F = mg = k_{eq}x$

Rearranging for the equivalent constant:

$k_{eq}=\frac{mg}{x}$

Plugging in our values:
$k_{eq}=\frac{(25kg)\left ( 10\frac{m}{s^2}\right )}{0.2m}$

$k_{eq} = 1.25\frac{kN}{m}$

$5k_1 = 1.25\frac{kN}{m}$

$k_1 = 250\frac{N}{m}$

Now we can use the expression for the potential energy stored in a spring:

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

The problem statement is asking for the total potential energy stored, so we need to use the total distance stretched (which includes both from the mass and manually). Plugging in our values, we get:

$U = \frac{1}{2}\left ( 250\frac{N}{m}\right )\left ( 1.2m\right )^2$

A spring of rest length $1\textup{ m}$ is used to hold up a $2.55*10^5\textup{ kg}$ rocket from the bottom as it is prepared for the launch pad. The spring compresses to $.775\textup{ m}$ .

Later, the same spring is used to support a rocket of mass $1.29*10^5\textup{ kg}$ . Determine the compression.

$.114\textup{ m}$

None of these

$.098\textup{ m}$

$.75\textup{ m}$

$.29\textup{ m}$

Explanation

The spring force is going to add to the gravitational force to equal zero.

Plugging in values:

Solving for

$k=1.11*10^7\frac{N}{m}$

Once again, using

Solving for

An object is attached to a spring, and is stretched 3m. If the restoring force is equal to , what is the spring constant?

$2100 \frac{N}{m}$

$-2100 \frac{N}{m}$

$700 \frac{N}{m}$

$-700 \frac{N}{m}$

Explanation

Hooke's law states that the spring force is equal to the product of the spring constant and the displacement of the spring:

$F_{s}=-kx$

The force is negative because it acts in the direction opposite of the displacement from the equilibrium position (i.e. when we stretch we do so in the positive direction). We are given the force and the displacement, so we just solve for k:

$\frac{-6300}{-3}=2100 \frac{N}{m}$

Two springs with unknown constants are attached in a linear fashion and hung from a ceiling. However, we do know that the constant of one spring is times that of the other. If a mass of is attached to the lower spring and the system stretches a total distance of , what is the constant of the weaker spring?

$g =10\frac{m}{s^2}$

$42.9\frac{N}{m}$

$34.7\frac{N}{m}$

$25.0\frac{N}{m}$

$35.0\frac{N}{m}$

$51.3\frac{N}{m}$

Explanation

Much like resistors in a circuit, spring constants attached in parallel and linear fashion can be combined to a single constant. Since the springs are attached in a linear fashion, we will use the following expression:

$\frac{1}{k_t}=\frac{1}{k_1}+\frac{1}{k_2}$

Which becomes:

$k_t = \frac{k_1k_2}{k_1+k_2}$

From the problem statement, we know that:

So let's substitute that into the problem:

$k_t = \frac{1.4k_1^2}{k_1+1.4k_1}$

$k_t = \frac{1.4k_t^2}{2.4k_t} = 0.58\bar{3}k_1$

Now we that we have a "total" spring constant, we can use Hooke's Law:

$k_t= \frac{mg}{x}$

Plugging in our values, we get:

$k_t = \frac{(10kg)\left ( 10\frac{m}{s^2}\right )}{4m}$

$k_t = 25\frac{N}{m}$

$25\frac{N}{m}= 0.58\bar 3 k_1$

$k_1 = 42.9\frac{N}{m}$

A narrow spring is placed inside a wider spring of the same length. The spring constant of the wider spring is twice that of the narrow spring. The two-spring-system is used to hold up a box of mass . They compress by .

Determine the spring constant of the narrower spring.

$1.63*10^5\frac{N}{m}$

$6.48*10^5\frac{N}{m}$

$1.13*10^5\frac{N}{m}$

$3.22*10^5\frac{N}{m}$

$4.91*10^5\frac{N}{m}$

Explanation

Use Hooke's law:

$F_{spring}=kx$

and

$F_1+F_2+...=F_{total}$

Wide spring:

$F_{ws}=k_1x$

Narrow spring:

$F_{ns}=k_2x$

From given information:

Substitute:

$F_{ws}=2k_2x$

Summation of forces:

Where is pointing down and thus a negative value. Convert to and plug in values:

Solve for :

$k_2=1.63*10^5\frac{N}{m}$

A spring is attached to the ceiling of an elevator with a block of mass $3\textup{ kg}$ hanging from it. If the displacement of the spring is $0.4\textup{ m}$ while the elevator is at rest, what is the displacement of the spring when the elevator begins accelerating upward at a rate of $2\ \frac{\textup{m}}{\textup{s}^2}$ .

$g=10\ \frac{\textup{m}}{\textup{s}^2}$

$0.48\textup{ m}$

$0.32\textup{ m}$

$0.08\textup{ m}$

$0.25\textup{ m}$

$0.55 \textup{ m}$

Explanation

When the elevator is at rest, we can use the following expression to determine the spring constant:

Where the force is simply the weight of the spring:

Rearranging for the constant:

$k = \frac{mg}{x}$

Now solving for the constant:

$k = \frac{(3kg)\left ( 10\frac{m}{s^2}\right )}{(0.4m)}$

$k = 75\frac{N}{m}$

Now applying the same equation for when the elevator is accelerating upward:

Where a is the acceleration due to gravity PLUS the acceleration of the elevator. Rearranging for the displacement:

$x = \frac{ma}{k}$

Plugging in our values:

$x = \frac{(3kg)\left ( 12\frac{m}{s^2}\right )}{\left ( 75\frac{N}{m}\right )}$

If you're confused why we added the acceleration of the elevator to the acceleration due to gravity. Think about the situation practically. When you are riding an elevator and it begins to accelerate upward, your body feels heavier. That's because your relative weight has increased due to the increased normal force due to a relative increase in acceleration.

Two springs with unknown constants are attached in a linear fashion and hung from a ceiling. However, we do know that the constant of one spring is twice that of the other. A mass of is attached to the bottom spring and the system drops . If you manually pull down the mass so the system stretches another , what is the maximum velocity of the block?

$2.7\frac{m}{s}$

$2.0\frac{m}{s}$

$1.3\frac{m}{s}$

$0.6\frac{m}{s}$

$3.4\frac{m}{s}$

Explanation

Since we are never asked about anything pertaining to an individual spring, their individual constants and arrangement are irrelevant. We can simply use an equivalent resistance to solve this problem. Beginning with Hooke's Law:

$F = mg = k_{eq}x$

Rearranging for the spring constant:

$k_{eq} = \frac{mg}{x}$

Plugging in our values, we get:

$k_{eq} = \frac{(8kg)\left ( 10\frac{m}{s^2}\right )}{0.8m}$

$k_{eq} = 100\frac{N}{m}$

Now we have a value for our equivalent resistance. At this point, the system has set into a new equilibrium. This is very important to remember when performing further calculations.

Let's move on to the next relevant information: we manually pull down the mass an additional distance. We are ultimately looking for the maximum velocity of the spring. This occurs when the mass passes through our new equilibrium. Furthermore, we have enough information to calculate the potential energy of the spring when it is at it's maximum stretched length. Therefore, we can use the expression for conservation of energy:

Where the initial condition is when the spring is full stretched, and the final condition is when the mass passes through equilibrium. Therefore, we can eliminate initial kinetic and final potential energy to get:

Plugging in expressions for each of these, we get:

$\frac{1}{2}k_{eq}x^2=\frac{1}{2}mv^2$

$k_{eq}x^2=mv^2$

Rearranging for velocity, we get:

$v = \sqrt{\frac{k_{eq}x^2}{m}}$

We have all of these values, so time to solve the problem:

$v = \sqrt{\frac{\left ( 100\frac{N}{m}\right )(0.6m)^2}{5kg}}$

$v = 2.7\frac{m}{s}$

Two springs are attached independently from a ceiling with the constant of one being $k_1 = 20\frac{N}{m}$ and the second constant being $k_2 = 30\frac{N}{m}$ . A third spring with a constant $k_3 = 40\frac{N}{m}$ is then attached to the bottom of both of the other springs. What is the total equivalent spring constant of this system?

$22.2\frac{N}{m}$

$14.5\frac{N}{m}$

$36.7\frac{N}{m}$

$45.8\frac{N}{m}$

$59.1\frac{N}{m}$

Explanation

Much like resistors in a circuit, spring constants can be combined to obtain a single equivalent constant. When springs are in parallel, we simply add their constants, and when they are added linearly, we add their inverses. Just like a circuit, let's begin with the two springs in parallel:

$k_{12}=k_1+k_2$

$k_{12} = 20\frac{N}{m}+30\frac{N}{m}$

$k_{12}=50\frac{N}{m}$

Now we can combine this equivalent constant with the third spring:

$\frac{1}{k_{123}}=\frac{1}{k_{12}}+\frac{1}{k_3}$

$k_{123} = \frac{k_{12}k_3}{k_{12}+k_3}$

$k_{123}= \frac{\left ( 50\frac{N}{m}\right )\left ( 40\frac{N}{m}\right )}{50\frac{N}{m}+40\frac{N}{m}}$

$k_{123}= 22.2\frac{N}{m}$

A spring is attached to a rope that is hanging from the ceiling. A block of mass is attached to the end of the spring which has a constant of $k = 50\frac{N}{m}$ . If the maximum tension felt in the rope is , what is the amplitude of the spring?