Physics, 09.09.2019 20:30 amakayla57

You are an industrial engineer with a shipping company. as part of the package- handling system, a small box with mass 1.60 kg is placed against a light spring that is compressed 0.280 m. the spring has force constant 48.0 n/m . the spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is μk = 0.300. when the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) what is the speed of the box at the instant when it leaves the spring? (b) what is the maximum speed of the box during its motion?

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Answer from: kaytek1

Elastic potential energy is the energy stored due to the deformation of an elastic object

(a) The speed at the instant the box leaves the spring is approximately 0.84 m/s

(b) The maximum speed of the box is approximately 0.996 m/s

Reason:

Known parameters are;

Mass of the box, m = 1.60 kg

The compression of the spring, x = 0.280 m

The force constant of the spring, K = 48.0 N/m

Coefficient of kinetic friction between the box and the horizontal surface, $\mu_k$ = 0.300

(a) The potential energy of the spring = 0.5 × K × d²

The normal reaction of the block on the surface, N = 1.60 kg × 9.81 m/s² = 15.696 N

The friction force, F_f = N × $\mu_k$

F_f = 15.696 N × 0.30 = 4.7088 N

Work done by friction, W_f = F_f × (d - x)

We get;

0.5 × K × d² - F_f × (d - x) = 0.5·m·v² + 0.5 × K × x²

Which gives;

28 × 0.28² - 4.7088×(0.28 - x) = 0.5 × 1.6 × v² + 24·x²

1.8816 - 1.318464 + 4.7088·x = 0.8·v² + 24·x²

0.563136 + 4.7088·x = 0.8·v² + 24·x²

0.563136 + 4.7088·x - 24·x² = 0.8·v²

Therefore, when x = 0, we have;

0.563136 + 4.7088*0 - 24*0² = 0.8·v²

Which gives;

0.563136 = 0.8·v²

$v = \sqrt{\dfrac{0.563136}{0.8} } \approx 0.84$

The speed at the instant the box leaves the spring, v ≈ 0.84 m/s

(b) The maximum speed is given as follows;

0.563136 + 4.7088·x - 24·x² = 0.8·v²

$v =\sqrt{ 0.70392+ 5.886 \cdot x - 30 \cdot x^2}$

At the maximum velocity, $v_{max}$ , we have;

$\dfrac{dv}{dx } = 0 = \dfrac{d}{dx} \left(\sqrt{ 0.70392+ 5.886 \cdot x - 30 \cdot x^2}\right) = \dfrac{5.886- 60 \cdot x}{2 \times \sqrt{ 0.70392+ 5.886 \cdot x - 30 \cdot x^2} }$

Which gives;

5.886 - 60·x = 0

5.886 - 60·x

x = 0.0981

$\therefore v_{max} =\sqrt{ 0.70392+ 5.886 \times 0.0981 - 30 \times 0.0981^2} \approx 0.996$

The maximum speed of the box during the motion, $v_{max}$ ≈ 0.996 m/s

Learn more about potential energy of spring here:

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Answer from: fatherbamboo

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

$W_s = \frac{1}{2}kc^2$

Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

You are an industrial engineer with a shipping company. as part of the package- handling system, a s

Answer from: Quest

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

Answer from: Quest

Answertechnician (a) is correct;

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