Physics, 06.11.2019 22:31 Tcareyoliver

Ahorizontal spring with a spring constant 100 n/m is compressed 20cm and used to launch a 2.5kg box across a frictionless, horizontalsurface. after the box travels some distance, the surface becomesrough. the coefficient of kinetic friction of the box on thesurface is 0.15. use work and energy to find how far the box slidesacross the rough surface before stopping.

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Answer from: jociymartinez6214

The distance traveled by the box along the rough surface before stopping is 0.54 m.

The given parameters;

spring constant, k = 100 N/mextension of the spring, x = 20 cm = 0.2mass of the box, m = 2.5 kgcoefficient of kinetic friction, μ = 0.15

The distance traveled by the box is determined by using work-energy theorem as shown below;

$\mu_k Fd = \frac{1}{2}kx^2\\\\d = \frac{kx^2}{2\mu_k F} \\\\d = \frac{kx^2}{2\mu_k \times mg} \\\\d = \frac{100(0.2^2)}{2\times 0.15 \times 2.5 \times 9.8} \\\\d = 0.54 \ m$

Thus, the distance traveled by the box along the rough surface before stopping is 0.54 m.

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Answer from: caarolin9

d = 0.544 m

Explanation:

To solve this problem we must work in two parts: one when the surface has no friction and the other when the surface has friction

Let's start with the part without rubbing, let's find the speed that the box reaches., For this we use the conservation of mechanical energy in two points: maximum compression and when the box is free (spring without compression)

Initial, maximum compression

Em₀ = Ke = ½ k x²

Final, free box without compressing the spring

$Em_{f}$ = K = ½ m v²

Emo = $Em_{f}$

½ k x² = ½ m v²

v = √ (k / m) x

Let's reduce the SI units measures

x = 20 cm (1m / 100cm) = 0.20 m

v = √ (100 / 2.5) 0.20

v = 1,265 m / s

Let's work the second part, where there is friction. In this part the work of the friction force is equal to the change of mechanical energy

$W_{fr}$ = ΔEm = $Em_{f}$ - Em₀

$W_{fr}$ = - fr d

Final point. Stopped box

$Em_{f}$ = 0

Starting point, starting the rough surface

Em₀ = K = ½ m v²

With Newton's second law we find the force of friction

fr = μ N

N-W = 0

N = W = mg

fr = μ mg

-μ m g d = 0 - ½ m v²

d = ½ v² / (μ g)

Let's calculate

d = ½ 1,265² / (0.15 9.8)

d = 0.544 m

Answer from: Quest

6/12=1/2

step-by-step explanation:

Answer from: Quest

It should be d if not then c

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Ahorizontal spring with a spring constant 100 n/m is compressed 20cm and used to launch a 2.5kg box...

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