A 2.4 kg breadbox on a frictionless incline of angle θ = 44 ˚ is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 100 N/m, as shown in the figure. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10.9 cm down the incline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction of the box's acceleration at the instant the box momentarily stops?

"Hooke's Law" states us the force that compress and expand a spring. From here on we can develop a relationship between the force applied and the elasticity of an object. In equation form "Hooke's law" is,

F = −kx

Where,

k is the spring constant and

x is the length of extension of a spring.

Given that,

m = 2.4 Kg

[tex]\theta=44^{\circ}[/tex]

(a) speed of the box:

To find for the speed of the box we can use the principle of conservation of energy which is equal to,

[tex]\mathrm{PE}_{\mathrm{b}}=\mathrm{PE}_{\mathrm{s}}+\mathrm{KE}_{\mathrm{b}}[/tex]

[tex]m g \times \sin \theta=\frac{1}{2} k x^{2}+\frac{1}{2} m v^{2}[/tex]

[tex]m g \times \sin \theta-\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}[/tex]

[tex]\frac{2 \mathrm{mg} \times \sin \theta-\mathrm{kx}^{2}}{m}=\mathrm{v}^{2}[/tex]

Substituting all of the known values,

[tex]\mathrm{v}^{2}=\frac{2 \times 2.4 \times 9.8 \times \sin 44^{\circ}-100 \times 0.109^{2}}{2.4}[/tex]

[tex]\mathrm{v}^{2}=\frac{32.67-1.1881}{2.4}[/tex]

[tex]\mathrm{v}^{2}=\frac{31.4819}{2.4}[/tex]

[tex]v^{2}=13.117[/tex]

[tex]\mathrm{v}=\sqrt{13.117}[/tex]

v = 3.62

v = 3.62 m/s

We can conclude that the box move at a speed of 3.62 m/s down to the 10.9 cm.

(b) Distance that traveled down from point of release:

We will use again the principle of conservation of energy but this time we will find the distance where the box stops. So since that point where the box will stop the kinetic energy of the box will become zero so our equation will be,

[tex]\mathrm{PE}_{\mathrm{b}}=\mathrm{PE}_{\mathrm{s}}[/tex]

[tex]\mathrm{mg} \times \sin \theta=\frac{1}{2} \mathrm{kx}^{2}[/tex]

[tex]\frac{2 \mathrm{mg} \sin \theta}{k}=\mathrm{x}[/tex]

Substituting all of the known values,

[tex]x=\frac{2 \times 2.4 \times 9.8 \times \sin 44^{\circ}}{100}[/tex]

[tex]x=\frac{32.67}{100}[/tex]

x = 0.3267 m

The distance where the box will momentarily stop is 0.3267 m.

(c) and (d) Acceleration and magnitude:

For the acceleration of the boxed we can use the Newton's second law,

[tex]\mathrm{a}=\frac{F_{n e t}}{m}[/tex]

Base on the diagram we can see that the force acting on the box is in the tension on the cord and the gravitational force. Remember that the positive force is going up the inclined and the negative force is going down the inclined, so we have,

[tex]\mathrm{F}_{\mathrm{net}}=\mathrm{T}-\mathrm{Fg}[/tex]

[tex]\mathrm{F}_{\mathrm{net}}=\mathrm{kx}-\mathrm{mg} \sin \theta[/tex]

[tex]\mathrm{F}_{\mathrm{net}}=(100 \mathrm{N} / \mathrm{m})(0.3267 \mathrm{m})-(2.4)\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right) \sin \left(44^{\circ}\right)[/tex]

[tex]\mathrm{F}_{\text {net }}=32.67-16.32[/tex]

[tex]\mathrm{F}_{\mathrm{net}}=16.34 \mathrm{N}[/tex]

We know that the acceleration,

[tex]\mathrm{a}=\frac{F_{n e t}}{m}[/tex]

[tex]a=\frac{16.34}{2.4}[/tex]

[tex]a=6.80 \mathrm{m} / \mathrm{s}^{2}[/tex]

We can conclude that the acceleration of the boxed when it instantly stop is

[tex]6.287 \mathrm{m} / \mathrm{s}^{2} \text { going up the inclined. }[/tex]