The diagram shows a mass, m = 6.70kg, at rest against a spring of force constant k = 1560 N/m,

Question

The diagram shows a mass, m = 6.70kg, at rest against a spring of force constant k = 1560 N/m, and compressed a distance Δx = 0.200m. When released, the mass will slide on a horizontal, rough surface with a coefficient of kinetic friction 𝜇_k = 0.220. How far will the mass travel on the rough surface before coming to rest?

DO NOT USE KINEMATICS - NO REFERENCE TO ACCELERATION.

Here is the diagram given, I need help calculating the distance the box travels, work and energy is a new topic for me so I have no idea what to do. If you can help me through the problem by showing me the steps along with an explanation it would really help, thank you!

Answer 1

I cannot see the diagram, so I will work from my imagination created by the description.

Let the x-axis be along the movement of the mass.

Let x = 0 be the point where the the spring is neither compressed nor stretched.

The mass will be oscillating between the positive and negative x positions until it stops moving.

Initially the mass is at the position

x₀ = -0.2m

After release it will move towards x = 0 and then continue moving, stretching the string.

At some point it reaches velocity 0 and starts its backward move.

To solve the problem you need the following

PROPERTY 1)

F = -kx

where

F is the force, which the spring exerts on the mass.

The sign is negative because the force always acts against the displacement:

When the mass is at a positive displacement x, the force is pulling it backward,

and vice versa.

PROPERTY 2)

The force of friction is constant in magnitude, but its direction is always against movement.

Let

g = 9.81 m/s² be gravitational acceleration,

μ = 0.22 be the coefficient of kinetic friction,

R be the force of friction.

The magnitude of the force of friction is

|R| = mgμ

When the mass is increasing its x-position then

R = -mgμ

and when the mass is decreasing its x position then

R = mgμ.

PROPERTY 3)

There are two forms of energy involved:

kinetic energy of the mass, which depends on its speed, and

spring energy E, which depends on the mass's displacement x. Specifically

E = kx²/2

Initially the mass is at rest, so its kinetic energy is 0.

And the spring energy originally is

E₀ = kx₀²/2.

As the mass starts moving, its kinetic energy increases at the expense of the

spring energy.

After the mass passes the point x = 0, its speed starts decreasing,

transferring its kinetic energy into the spring energy, which then keep increasing.

PROPERTY 4)

This back-and-forth transfer of energy between kinetic and spring energy would

continue forever, if it were not for friction, which keeps dissipating some of the energy.

The amount of dissipated energy is always equal the "work" performed by the

force of friction. (This is the work-energy theorem.)

In general, work done by force over a distance is defined as the product of the force and the distance.

(This is the formulation for the special case of a constant force acting in parallel to the distance, as is the case in your situation.)

Now we use the above properties to solve the problem.

I do not know whether in the statement of the problem the phrase "coming to rest" refers to

(A) the first time when the mass's velocity is 0, or

(B) the time the whole oscillation stops.

So I will show you how to solve the problem for both interpretations.

INTERPRETATION (A):

The first time the mass comes to a stop will be at some point x₁.

We want to calculate x₁.

By the time it got to x₁, it travelled a distance (x₁ - x₀).

At the point x₁ the mass has speed 0, so has kinetic energy 0, so

all the initial spring energy E₀ got converted to the new spring energy

E₁ = kx₁²/2

plus the work of friction, which is

W = |R|(x₁ - x₀)

So the condition for reaching velocity of 0 for the first time is

E₀ = E₁ + W

Substituting into the above:

kx₀²/2 = kx₁²/2 + mgμ(x₁ - x₀)

This can be manipulated as follows

kx₁²/2 - kx₀²/2 + mgμ(x₁ - x₀) = 0

k(x₁ + x₀)(x₁ - x₀)/2 + mgμ(x₁ - x₀) = 0

Discarding the uninteresting trivial solution x₁ = x₀:

k(x₁ + x₀)/2 + mgμ = 0

x₁ = -x₀ - 2mgμ/k

After substituting actual numbers

x₁ = 0.2 - 2×6.7×9.81×0.22/1560 = 0.18

So the mass reaches speed of 0 for the first time at the point x₁ = 0.18,

by which time it will have travelled a distance of 0.38 m.

INTERPRETATION (B):

After oscillating for some time the mass comes to its final resting place

at some position x_f, having travelled some total distance X.

We want to calculate X.

At the the final place all the initial spring energy E₀ got converted to

the spring energy at the final place

E_f = kx_f²/2

plus the work performed by the force of friction over that total distance X

W_f = |R|X

So the conservation of energy is expressed by

E₀ = E_f + W_f

Substituting into the above:

kx₀²/2 = kx_f²/2 + mgμX

From that

X = (kx₀²/2 - kx_f²/2)/mgμ = k(x²₀ - x_f²)/2mgμ (1)

It remains to calculate x_f.

At the final resting place the total force acting on the mass is 0, i.e.,

R + F = 0

Substituting for F:

R - kx_f = 0

From that

x_f = R/k

which could be either positive or negative.

(It is difficult to predict on which side of x=0 the oscillation will stop).

But we do not need to know the sign of x_f, because we just need

x_f²= (R/k)² = (mgμ/k)²

Substituting actual numbers

x_f² = (6.7×9.81×0.22/1560)² = 0.000086 m²

Now we can substitute actual numbers into (1)

X = 1560×(0.2² - 0.000086)/(2×6.7×9.81×0.22) = 2.153 m

So the mass stops oscillating after it travelled a total distance of approximately 2.153 m.