# Time a spring takes to slow a mass

• LANS
In summary, the mass travels 0.1705 meters until it stops. It takes 1/4 of a second for the mass to travel this distance.
LANS

## Homework Statement

A mass M is attached to a spring with spring constant K. At the equilibrium point of the spring, the mass has a velocity of V.
M = 8.07 kg
K = 113 N/m
V_o = 0.638 m/s

How far does the mass travel until it stops? How long (in seconds) does it take for the mass to travel from the equilibrium point until it stops?

## Homework Equations

$$\frac{1}{2}MV_o^2 = \frac{1}{2}Kx^2$$ - equation 1

$$F(x) = kx$$ - spring force

$$V(x) = \sqrt{V_o^2 - \frac{Kx^2}{m}}$$ - from energy.

## The Attempt at a Solution

Using equation 1, I can solve part 1 easily. I plug in M,K,V to equation 1and solve for x, which gives me $$x = 0.1705m$$

I have no idea how to solve part 2. I've tried using power, but that doesn't go anywhere meaningful.
$$P(x) = F(x)*V(x)$$

Integrating for total power gives me
$$\frac{MV_o^2}{2t} = \int F(x)*V(x)$$

Simplifying the integral:
$$\frac{MV_o^2}{2t} = \int \sqrt{K^2 x^2 V_o^2 - \frac{K^3 x^4}{m}}$$

I've tried solving that for t, and it doesn't give me the right answer. I haven't thought of it yet, but I feel like there should be an easier solution to this problem. Any help is appreciated.

It might be easier to cast this as a SHO formulation where T can be found directly from the other variables. in other words can you convert this to the form

x(t)=A sin (wt) and solve for T. hint: w=sqrt(k/m) and the time to stop is 0.5T.

Have you learned the formula for the period of a mass-spring system? The time it takes to go from equilibrium to rest is actually 1/4T (not 0.5T).

ideasrule said:
Have you learned the formula for the period of a mass-spring system? The time it takes to go from equilibrium to rest is actually 1/4T (not 0.5T).

my bad. I forgot it was a quarter cycle and not 1/2, thanks for the correction.

## What is the equation for calculating the time a spring takes to slow a mass?

The equation for calculating the time a spring takes to slow a mass is:
t = 2π√(m/k)
where t is the time in seconds, m is the mass in kilograms, and k is the spring constant in newtons per meter.

## How does the mass affect the time it takes for the spring to slow it down?

The mass directly affects the time it takes for the spring to slow it down. The larger the mass, the longer it will take for the spring to slow it down. This is because a larger mass requires more force to slow down, and the spring will need more time to exert that force.

## What is the spring constant and how does it impact the time of slowing down?

The spring constant is a measure of the stiffness of a spring. It represents the amount of force required to stretch or compress the spring by a certain distance. The higher the spring constant, the faster the spring will slow down the mass. This is because a higher spring constant means the spring can exert more force in a shorter amount of time.

## Does the length of the spring affect the time it takes to slow down a mass?

Yes, the length of the spring does affect the time it takes to slow down a mass. A longer spring will have a longer distance to travel to slow down the mass, so it will take more time. On the other hand, a shorter spring will have a shorter distance to travel and will take less time to slow down the mass.

## Are there any other factors that can impact the time a spring takes to slow a mass?

Yes, there are some other factors that can impact the time a spring takes to slow a mass. These include the initial velocity of the mass, air resistance, and the presence of any other external forces. These factors can complicate the calculation, but the basic equation still holds true for determining the time a spring takes to slow a mass.

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